/</,  /\  Z  3 


University  of  California  •  Berkeley 

Gift  of 
MRS.  GRIFFITH  C.  EVANS 


ELEMENTS 


OP 


DESCRIPTIVE    GEOMETRY, 


WITH  ITS  APPLICATIONS  TO 


SPHERICAL  PROJECTIONS,   SHADES  AND   SHADOWS* 
PERSPECTIVE  AND  ISOMETRIC  PROJECTIONS. 


BY 
ALBERT  E.  CHURCH,  LL.  D., 

PROFESSOR  OP  MATHEMATICS  IN  THE  U.  S.  MILITARY  ACADEMY,  AUTHOR  OP 

ELEMENTS  OF  THE  DIFFERENTIAL  AND  INTEGRAL  CALCULUS, 

AND  OF  ELEMENTS  OF  ANALYTICAL  GEOMETRY. 


A.  S.  BAKJNES  &  COMPANY, 

NEW  YORK  AND  CHICAGO. 
1872. 


Valuable  forks  Hj  Leading  Autbors 

IN  THB 

HIGHER  MATHEMATICS* 


A..    E.    CHURCH,    LL.D., 

Prof.  Mathematics  in  the   U.  S.  Military  Acadenty,  West  Point, 
CHURCH'S    ANALYTICAL    GEOMETRY. 
CHURCH'S    CALCULUS. 
CHURCH'S    DESCRIPTIVE    GEOMETRY 


EDWARD    3Z.    COURTENAY,    LL.D., 

Late  Prof.  Mathematics  in  the  University  of  Virginia, 
COURTENAY'S    CALCULUS. 


CKLAS.   W.    HACKLEY,    S.  T.  r>.9 

Late  Prof,  of  Mathematics  and  Astronomy  in  Columbia  College, 
HACKLEY'S    TRIGONOMETRY. 


W.    H.    CBARTLETT,    XJL,.r>., 

Prof.  ofJVat.  &  Exp.  Philos.  in  the  U.  S.  Military  Acad.,  West  Point, 
BARTLETT'S    SYNTHETIC    MECHANICS. 
BARTLETTS    ANALYTICAL    MECHANICS. 
BARTLETT'S    ACOUSTICS    AND    OPTICS, 
BARTLETT'S    ASTRONOMY. 


DAT1ES    «&    PECK, 

.  Department  of  Mathematics,  Columbia  College, 
MATHEMATICAL    DICTIONARY. 


CHARLES   DAVIES,    LL.D., 

Late  of  the  United  States  Military  Academy  and  of  Columbia  College, 

A    COMPLETE    COURSE    IN    MATHEMATICS. 

See  A.  S.  Babnes  &  Co.'s  Descriptive  Catalogue. 


Entered,  according  to  Act  of  Congress,  in  the  year  1864,  by 
BARNES   &   BURR, 
In  the  Clerk's  Office  of  the  District  Court  of  the  United  States  for  the  Southern  District  of 
New  York. 


PREFACE. 


These  pages  have  been  written  and  are  published  with 
the  single  object  of  presenting,  in  proper  form  to  be  used 
as  a  text-book,  the  course  of  Descriptive  Geometry,  as 
taught  at  the  U.  S.  Military  Academy. 

Without  any  effort  to  enlarge  or  originate,  the  author 
has  striven  to  give,  with  a  natural  arrangement  and  in 
clear  and  concise  language,  the  elementary  principles  and 
propositions  of  this  branch  of  science,  of  so  much  interest 
to  the  mathematical  student,  and  so  necessary  to  both  the 
civil  and  military  engineer. 

Though  indebted  for  many  of  the  ideas  to  the  early  in* 
structions  of  his  predecessor  and  friend,  Professor  Davies, 
whose  text-books  on  this  subject  were  among  the  first  in 
the  English  language,  the  author  has  been  much  aided  by 
a  frequent  reference  to  the  French  works  of  Leroy  and 
Olivier,  and  to  the  elaborate  American  work  of  Professor 
Warren. 

It  is  intended  to  include,  in  an  edition  to  be  issued  at 
an  early  day,  the  application  of  the  subject  to  shades  and 
shadows,  and  perspective. 

U.  S.  Military  Academy,  ) 
October,  1864  \ 


CONTENTS. 


PAET  I. 

ORTHOGRAPHIC  PROJECTION. 

TJMU 

Preliminary  definitions 1 

Representation  of  points 2 

Representation  of  planes . 4 

Representation  of  right  lines 5 

Revolution  of  objects 7 

Revolution  of  the  vertical  plane 8 

Notation  used  in  the  description  of  drawings 12 

Manner  of  delineating  the  different  lines  used 12 

Elementary  problems  relating  to  the  right  line  and  plane 13- 

Construction  and  classification  of  lines 33 

Projection  of  curves 36 

Tangents  and  normals  to  lines 37 

Generation  and  properties  of  the  helix 39 

Generation  and  classification  of  surfaces 40 

Generation  and  properties  of  cylindrical  surfaces 42 

Generation  and  properties  of  conical  surfaces 44 

Warped  surfaces  with  a  plane  directer 47 

Generation  and  properties  of  the  hyberbolic  paraboloid 49 

Warped  surfaces  with  three  linear  directrices 51 

Generation  and  properties  of  the  helicoid 53 

Surfaces  of  revolution 54 

The  hyperboloid  of  revolution  of  one  nappe 56 

Tangent  planes  and  surfaces  ;  normal  lines  and  planes 61 

Problems  relating  to  tangent  planes  to  single  curved  surfaces 66 

Problems  relating  to  tangent  planes  to  warped  surfaces 74 

Problems  relating  to  tangent  planes  to  double  curved  surfaces 78 

Points  in  which  surfaces  are  pierced  by  lines 85 

Intersection  of  surfaces  by  planes.    Development  of  single  curved 

surfaces 86 


VI  CONTENTS. 

Intersection  of  curved  surfaces 09 

Development  of  an  oblique  cone 104 

Practical  problems 107 


PART  II. 

SPHERICAL  PROJECTIONS. 

Preliminary  definitions 113 

Orthographic  projections  of  the  sphere 116 

Stereographic  projections  of  the  sphere 121 

Globular  projections 130 

Gnomonic  projection 131 

Cylindrical  projection .  .• 133 

Conic  projection 133 

Construction  of  maps. •. . .  134 

Lorgna's  map 135 

Mercator's  chart 135 

Flamstead's  method 137 

The  Polyconic  method 138 

PART  III. 

SHADES  AND  SHADOWS. 

Preliminary  definitions 140 

Shadows  of  points  and  lines ' 143 

Brilliant  points 145 

Practical  problems 1 &e 

PAET  IY. 

LINEAR  PERSPECTIVE. 

Preliminary  definitions  and  principles 157 

Perspectives  of  points  andright  lines.    Vanishing  points  of  rightlines.  158 

Perspectives  of  curves 163 

Line  of  apparent  contour 163 


CONTENTS.  .  Vll 

PAGE 

Vanishing  points  of  rays  of  light  and  of  projections  of  rays 164 

Perspectives  of  the  shadows  of  points  and  right  lines  on  planes 165 

Practical  problems 167 

PAET  V. 

ISOMETRIC  PROJECTIONS. 

Preliminary  definitions  and  principles 186 

Isometric  projections  of  points  and  lines. .»»»., 188 

Practical  problems. .......   189 


PART  I. 

ORTHOGRAPHIC  PROJECTIONS. 


PRELIMINARY    DEFINITIONS. 

1.  Descriptive  Geometry  is  that  branch  of  Mathematics 
which  has  for  its  object  the  explanation  of  the  methods  of  rep- 
resenting by  drawings : 

First.  All  geometrical  magnitudes. 

Second.  The  solution  of  problems  relating  to  these  magnitudes 
in  space. 

These  drawings  are  so  made  as  to  present  to  the  eye,  situated 
at  a  particular  point,  the  same  appearance  as  the  magnitude  or 
object  itself,  were  it  placed  in  the  proper  position. 

The  representations  thus  made  are  the  projections  of  the  mag- 
nitude or  object. 

The  planes  upon  which  these  projections  are  usually  made  are 
the  planes  of  projection. 

The  point,  at  wliich  the  eye  is  situated,  is  the  point  of  sight. 


2.  When  the  point  of  sight  is  in  a  perpendicular,  drawn  to 
the  plane  of  projection,  through  any  point  of  the  drawing,  and 
at  an  infinite  distance  from  this  plane,  the  projections  are  Ortho- 
graphic. 


2  DESCRIPTIVE   GEOMETRY. 

When  the  point  of  sight  is  within  a  finite  distance  of  the 
drawing,  the  projections  are  Scenographic,  commonly  called  the 
Perspective  of  the  magnitude  or  object. 

3.  It  is  manifest  that,  if  a  straight  line  be  drawn  through  a 
given  point  and  the  point  of  sight,  the  point,  in  which  this  line 
pierces  the  plane  of  projection,  will  present  to  the  eye  the  same 
appearance  as  the  point  itself,  and  therefore  be  the  projection  oj 
the  point  on  this  plane. 

The  line  thus  drawn  is  the  projecting  line  of  the  point, 

4.  In  the  Orthographic  Projection,  since  the  point  of  sight 
is  at  an  infinite  distance,  the  projecting  lines  drawn  from  any 
points  of  an  object,  of  finite  magnitude,  to  this  point,  will  be 
parallel  to  each  other  and  perpendicular  to  the  plane  of  projection. 

In  this  projection  two  planes,  are  used,  at  right  angles  to  each 
other,  the  one  horizontal  and  the  other  vertical,  called  respect- 
ively the  horizontal  and  vertical  plane  of  projection. 

5.  In  Fig.  1,  let  the  planes  represented  by  ABF'  and  BAD 
be  the  two  planes  of  projection,  the  first  the  horizontal  and  the 
second  the  vertical. 

Their  line  of  intersection  AB  is  the  ground  line. 

These  planes  form  by  their  intersection  four  diedral  angles. 
The  first  angle,  in  which  the  point  of  sight  is  always  situated, 
is  above  the  horizontal  and  in  front  of  the  vertical  plane.  The 
second  is  above  the  horizontal  and  behind  the  vertical.  The 
third  is  below  the  horizontal  and  behind  the  vertical.  The 
fourth,  below  the  horizontal  and  in  front  of  the  vertical,  as 
marked  in  the  figure*. 

REPRESENTATION     OP    POINTS. 

6.  Let  M,  Fig.   1,  be   any  point  in  space.     Through  it  draw 


DESCRIPTIVE   GEOMETRY.  3 

Mm  perpendicular  to  the  horizontal,  and  Mm'  perpendiculai 
to  the  vertical  plane ;  m  will  be  the  projection  of  M  on  tty? 
horizontal,  and  m'  that  on  the  vertical  plane,  Art.  (4).  Hence, 
the  horizontal  projection  of  a  point  is  the  foot  of  a  perpendicular 
through  the  point,  to  the  horizontal  plane ;  and  the  vertical  pro- 
jection of  a  point  is  the  foot  of  a  perpendicular  through  it  to  the 
vertical  plane. 

The  lines  Mm  and  Mm'  are  the  horizontal  and  vertical  pro- 
jecting lines  of  the  point. 

I 

V.  Through  the  lines  *2m  and  Mm'  pass  a  plane.  It  will  be 
perpendicular  to  both  planes  of  projection,  since  it  contains  a 
right  line  perpendicular  to  each,  and  therefore  perpendicular 
to  the  ground  line  AB.  It  intersects  these  planes  in  the  lines 
mo  and  mro1  both  perpendicular  to  AB  at  the  same  point,  form- 
ing the  rectangle  Mo.  By  an  inspection  of  the  figure  it  is  seen 
that 

Mm  =  m'o,     and     Mm'  =  mo ; 

that  is,  the  distance  of  the  point  M,  from  the  horizontal  plane, 
is  equal  to  the  distance  of  its  vertical  projection  from  the  ground 
line ;  and  the  distance  of  the  point  from  the  vertical  plane  is 
equal  to  that  of  its  horizontal  projection  from  the  ground  line. 

8.  If  the  two  projections  of  a  point  are  given,  the  point  is  com- 
p>letely  determined;  for  if  at  the  horizontal  projection  ?ny  a 
perpendicular  be  erected  to  the  horizontal  plane,  it  will  contain 
the  point  M.  A  perpendicular  to  the  vertical  plane  at  m\  will  also 
contain  M ;  hence  the  point  M  is  determined  by  the  intersectior 
of  these  two  perpendiculars. 

If  M  be  in  the  horizontal  plane,  Mm  =  0,  and  the  point  is  its 
own  horizontal  projection.  The  vertical  projection  will  be  in  the 
ground  line  at  o. 


DESCRIPTIVE    GEOMETRY. 


If  M  be  in  the  vertical  plane,  it  will  be  its  own  vertical  projec- 
tion, and  its  horizontal  projection  will  be  in  the  ground  line  at  o. 

If  M  be  in  the  gsyund  line,  it  will  be  its  own  horizontal,  and 
also  its  own  vertical  projection. 


REPRESENTATION    OF    PLANES. 

9.  Let  tTt',  Fig.  2,  he  a  plane,  oblique  to  the  ground  line, 
intersecting  the  planes  of  projection  in  the  lines  tT  and  t'T  re- 
spectively. It  will  be  completely  determined  in  position  by  its 
two  lines  tT  and  t'T, 

Its  intersection  with  the  horizontal  plane  is  the  horizontal  trace 
of  the  plane,  and  its  intersection  with  the  vertical  plane  is  the 
vertical  trace.     Hence,  a  plane  is  given  by  its  traces. 

Neither  trace  of  this  plane  can  be  parallel  to  the  ground  line ; 
for  if  it  should  be,  the  plane  would  be  parallel  to  the  ground 
line,  which  is  contrary  to  the  hypothesis. 

The  two  traces  must  intersect  the  ground  line  at  the  same 
point,  for  if  they  intersect  it  at  different  points,  the  plane  would 
intersect  it  in  two  points,  which  is  impossible. 

If  the  plane  be  parallel  to  the  ground  line,  as  in  the  same 
figure,  its  traces  must  be  parallel  to  the  ground  line ;  for  if  they 
are  not  parallel  th.ey.must  intersect  it;  in  which  case  the  plane 
would  have  at  least  one  point  in  common  with  the  ground  line, 
which  is  contrary  to  the  hypothesis. 

If  the  plane  be  parallel  to  either  vplane  of  projection,  it  will 
have  but  one  trace,  which  will  be  on  the  other  plane  and  par- 
allel to  the  ground  line. 

10.  If  the  given  plane  be  perpendicular  to  the  horizontal 
plane,  its  vertical  trace  will  be  perpendicular  to  the  ground  line, 
as  t'T,  in  Fig.  2;  for  the  vertical  plane  is  also  perpendiculai 
to  the  horizontal  plane.;  hence  the  intersection  of  the  two  planes 
which  is  the  vertical  trace,  must  be  perpendicular  to  the  hori 


DESCRIPTIVE    GEOMETRY.  C 

rontal  plane,  and  therefore  to  the  ground  line  which  passes 
through  its  foot. 

Likewise,  if  a  plane  be  perpendicular  to  the  vertical  plane, 
its  horizontal  trace  will  be  perpendicular  to  the  ground  line. 

If  the  plane  simply  pass  through  the  ground  line,  its  position 
is  not  determined. 

If  two  planes  are  parallel,  their  traces  on  the  same  plane  of 
projection  are  parallel,  for  these  traces  are  the  intersections  of 
the  parallel  planes  by  a  third  plane. 


REPRESENTATION    OF    RIGHT    LINES. 

11.  Let  MN,  Fig.  3,  be  any  right  line  in  space.  Through 
it  pass  a  plane  Mmn  perpendicular  to  the  horizontal  plane ; 
mn  will  be  its  horizontal,  and  pp\  perpendicular  to  AB,  Art. 
(10),  its  vertical  trace.  Also  through  MN  pass  a  plane  Mw'»', 
perpendicular  to  the  vertical  plane :  m'n'  will  be  its  vertical,  and 
o'o  its  horizontal  trace.  The  traces  mn  and  m'n'  are  the  pro- 
jections of  the  line.  Hence,  the  horizontal  projection  of  a  right 
line  is  the  horizontal  trace  of  a  plane  passed  through  the  line 
perpendicular  to  the  horizontal  plane ;  and  the  vertical  projection 
of  a  right  line  is  the  vertical  trace  of  a  plane,  through  the  line 
perpendicular  to  the  vertical  plane. 

The  planes  Mmn  and  Mm'ft'  are  respectively  the  horizontal 
and  vertical  projecting  planes  of  the  line. 

12.  The  two  projections  of  the  line  being  given,  the  line  will, 
in  general,  be  completely  determined;  for  if  through  the  hori- 
zontal projection  we  pass  a  plane  perpendicular  to  the  horizon- 
tal plane,  it  will  contain  the  line;  and  if  through  the  vertical 
projection  we  pass  a  plane  perpendicular  to  the  vertical  plane, 
it  will  also  contain  the  line.  The  intersection  of  these  planes 
must,  therefore,  be  the  line.  Hence  we  say,  a  right  line  is  given  j 
by  its  jirojections. 


6  DESCRIPTIVE    GEOMETRY. 

13.  The  projections  mn  and  m'n'  are  also  manifestly  made  up 
of  the  projections  of  all  the  points  of  the  line  MN.  Hence,  if  a 
right  line  pass  through  a  point  in  space,  its  projections  will 
vass  through  the  projections  of  the  point.  Likewise,  if  any  two 
points  in  space  be  joined  by  a  right  line,  the  projections  of  this 
line  will  be  the  right  lines  joining  the  projections  of  the  points  on 
the  same  plane. 


14.  If  the  right  line  be  perpendicular  to  eitlier  plane  of  pro- 
jection, its  projection  on  that  plane  will  be  a  point,  and  its  pro- 
jection on  the  other  plane  will  be  perpendicular  to  the  ground 
line.  Thus  in  Fig.  4,  Mm  is  perpendicular  to  the  horizontal 
plane:  m  is  its  horizontal,  and  m'o  its  vertical  projection. 

If  the  line  be  parallel  to  either  plane  of  projection,  its  pro- 
jection on  that  plane  will  evidently  be  parallel  to  the  line  itself 
and  its  projection  on  the  other  plane  will  be  parallel  to  the 
ground  line.  For  the  plane  which  projects  it  on  the  second 
plane  must  be  parallel  to  the  first;  its  trace  must  therefore  be 
parallel  to  the  ground  line,  Art.  (9);  but  this  trace  is  the  pro- 
jection of  the  line.  Thus  MN  is  parallel  to  the  horizontal 
plane,  and  m'n'  is  parallel  to  AB.  Also,  the  definite  portion 
MN,  of  the  line,  is  equal  to  its  projection  mn,  since  they  are 
opposite  sides  of  the  rectangle  Mn. 

If  the  line  is  parallel  to  both  planes  of  projection,  or  to  the 
ground  line,  both  projections  will  be  parallel  to  the  ground 
line. 

If  the  line  lie  in  either  plane  of  projection,  its  projection  on 
that  plane  will  be  the  line  itself,  and  its  projection  on  the 
other  plane  will  be  the  ground  line.  Thus  in  Fig.  5,  MN,  in 
the  vertical  plane,  is  its  own  vertical  projection,  and  mn  or  AB 
is  its  horizontal  projection. 

15.  If  the  two  projections  of  a  right  line  are  perpendicular 
to  the  ground  line,  the  line  is  undetermined,  as  the  two  pro- 


DESCRIPTIVE   GEOMETRY.  7 

jecting  planes  coincide,  forming  only  one  plane,  and  do  not  by 
their  intersection  determine  the  line  as  in  Art.  (12). 

All  lines  in  this  plane  will  have  the  same  projections.  Thus 
in  Fig.  5,  mn  and  m'n'  are  both  perpendicular  to  AB;  and 
anv  line  in  the  plane  MNo  will  have  these  for  its  projections. 

If,  however,  the  projections  of  two  points  of  the  line  are 
given,  the  line  will  then  be  determined;  that  is,  if  mm'  and 
nn'  are  given,  the  two  points  M  and  N  will  be  determined, 
and,  of  course,  the  right  line  which  joins  them. 

All  lines  and  points,  situated  in  a  plane  perpendicular  to 
either  plane  of  projection,  will  be  projected  on  this  plane  in 
the  corresponding  trace  of  the  plane. 


16.  If  two  right  lines  are  parallel,  their  projections  on  the  same 
plane  will  be  parallel.  For  their  projecting  planes  are  parallel, 
since  they  contain  parallel  lines  and  are  perpendicular  to  the 
same  plane;  hence  their  traces  will  be  parallel,  Art,  (10);  but 
these  traces  are  the  projections 


REVOLUTION    OF    OBJECTS* 

17.  Any  geometrical  magnitude  or  object  is  said  to  be  re- 
volved about  a  right  line  as  an  axis,  when  it  is  so  moved  that 
mch  of  its  points  describes  the  circumference  of  a  circle  whose 
plane  is  perpendicular  to  the  axis,  and  whose  centre  is  in  the  axis. 

By  this  revolution,  it  is  evident  that  the  relative  position  of 
the  points  of  the  object  is  not  changed,  each  point  remaining  at 
the  same  distance  from  any  of  the  other  points.  Thus,  if  the 
point  M,  Fig.  6,  be  revolved  about  an  axis  DD',  in  the  horizon- 
tal plane,  it  will  describe  the  circumference  of  a  circle  whose 
centre  is  at  o  and  whose  radius  is  Mo;  and  since  the  point 
must  remain  in  the  plane  perpendicular  to  DD',  when  it 
reaches  the  horizontal  plane  it  will  be  at  p  or  p\  in  the  per- 
pendicular mop,  at  a  distance  from  o  equal  to  Mo;  that  is,  it 


O  DESCRIPTIVE   GEOMETRY. 

will  be  found  in  a  straight  line  passing  through  its  horizontal 
projection  perpendicular  to  the  axis,  and  at  a  distance  from  the 
axis  equal  to  the  hypothenuse  of  a  right-angled  triangle  of  which 
the  base  (mo)  is  the  distance  from  the  horizontal  projection  to  the 
axis,  and  the  altitude  (Mm)  equal  to  the  distance  of  the  point 
from  the  horizontal  plane,  or  equal  to  the  distance  (mV)  of  its 
vertical  projection  from  the  ground  line. 

Likewise,  if  a  point  be  revolved  about  an  axis  in  the  verti- 
cal plane  until  it  reaches  the  vertical  plane,  its  revolved  position 
will  be  found  by  the  same  rule,  changing  the  word  horizontal 
into  vertical,  and  the  reverse. 

If  the  axis  pass  through  the  horizontal  projection  of  the 
point,  in  the  first  case,  the  base  of  the  triangle  will  be  0;  the 
hypothenuse  becomes  equal  to  the  altitude,  and  the  distance  to 
be  laid  off  will  be  simply  the  distance  from  the  vertical  pro- 
jection of  the  point  to  the  ground  line. 

In  like  manner,  its  revolved  position  will  be  found  when,  in 
the  second  case,  the  axis  passes  through  the  vertical  projection 
of  the  point. 

REVOLUTION    OF   THE   VERTICAL    PLANE, 

18.  In  order  to  represent  both  projections  of  an  object  on 
the  same  sheet  of  paper  or  plane,  after  the  projections  are  made 
as  in  the  preceding  articles,  the  vertical  plane  is  revolved  about 
the  ground  line  as  an  axis  until  it  coincides  with  the  horizontal 
plane,  that  portion  of  it  which  is  above  the  ground  line  falling 
beyond  it,  in  the  position  ABC'D',  Fig.  1,  and  that  part  which 
is  below  coming  up  in  front,  in  the  position  ABF'E'. 

In  this  new  position  of  the  planes  it  will  be  observed,  that 
the  planes  being  regarded  as  indefinite  in  extent,  all  that  part 
of  the  plane  of  the  paper  which  is  in  front  of  the  ground  line 
will  represent  not  only  that  part  of  the  horizontal  plane  which 
is  in  front  of  the  ground  line,  but  also  that  part  of  the  vertical 
plane  which  is  below   the  horizontal  plane;  while  the  part  be- 


DESCRIPTIVE    GEOMETRY.  9 

yond  the  ground  line  represents  that  part  of  the  horizontal  plane 
which  is  behind  the  vertical  plane,  and  also  that  part  of  the  verti- 
cal plane  which  is  above  the  horizontal  plane. 


19.  After  the  vertical  plane  is  revolved  as  in  the  preceding  ar- 
ticle, the  point  m',  in  Fig.  1,  will  take  the  position  m"  in  the 
line  mo  produced,  and  the  two  projections  m  and  m'  will  then  be 
in  the  same  straight  line,  perpendicular  to  AB.  Hence,  in  every 
drawing  thus  made,  the  two  projections  of  the  same  point  must  be 
in  the  same  straight  line,  perpendicular  to  the  ground  line. 

If,  then,  AB,  Fig.  7,  be  the  ground  line,  and  it  be  required  to 
represent  or  assume  a  point  in  space,  we  first  assume  m  for  its 
horizontal  projection;  through  m  erect  a  perpendicular  to  AB, 
and  assume  any  point,  as  m'  on  this  perpendicular,  for  the  vertical 
projection.  The  point  will  then  be  fully  determined,  Art.  (8). 
The  point  thus  assumed  is  in  the  first  angle,  above  the  horizontal 
plane  at  a  distance  equal  to  m'o,  and  in  front  of  the  vertical  plane 
at  a  distance  equal  to  mo,  Art.  (7). 

If  the  point  be  in  the  second  angle,  its  horizontal  projection  m 
must  be  on  that  part  of  the  horizontal  plane  beyond  the  ground 
line,  and  its  vertical  projection  m'  on  that  part  of  the  vertical 
plane  above  the  ground  line,  Art  (o).  When  the  latter  plane 
is  revolved  to  its  proper  position,  m'  will  fall  into  that  part  of  the 
horizontal  plane  beyond  the  ground  line,  and  the  two  projections 
will  be  as  in  (2),  Fig.  7,  mo  representing  the  distance  of  the 
point  behind  the  vertical  plane,  and  m'o  its  distance  above  the 
horizontal  plane. 

If  the  point  be  in  the  third  angle,  its  horizontal  projection  will 
be  be3Tond  the  ground  line,  and  its  vertical  projection  on  the  part 
of  the  vertical  plane  below  the  horizontal  plane.  The  vertical 
plane  being  revolved  to  its  proper  position,  m'  comes  in  front  of 
AB,  and  the  two  projections  will  be  as  in  (3).  t, 

If  the  point  be  in  the  fourth  angle,  the  two  projections  will  be 
as  in  (4). 


10  DESCRIPTIVE   GEOMETRY. 

20.  To  represent,  or  assume  a  plane  in  space,  we  draw  at  pleas- 
ure any  straight  line,  as  fT,  Fig.  8,  to  represent  its  horizontal 
trace ;  then  through  T,  draw  any  other  straight  line,  as  Tt't  to  rep- 
resent its  vertical  trace.  It  is  absolutely  necessary  that  these 
traces  intersect  AB  at  the  same  point,  if  either  intersects  it, 
Art.  (9). 

The  plane  and  traces  being  indefinite  in  extent,  the  portion  in- 
cluded in  the  first  angle  is  represented  by  tTt\  The  portion  in 
the  second  angle  by  t'Tt".  That  in  the  third  by  t''Tt'".  That 
in  the  fourth  by  t'"Tt. 

If  the  plane  be  parallel  to  the  ground  line  and  not  parallel  to 
either  plane  of  projection,  both  traces  must  be  assumed  parallel  to 
AB,  as  in  Fig.  9 ;  Tt  being  the  horizontal,  and  T*'  the  vertical 
trace,  Art.  (9). 

If  the  plane  be  parallel  to  either  plane  of  projection,  the  trace 
on  the  other  plane  is  alone  assumed,  and  that  parallel  to  AB. 

If  the  plane,  be  perpendicular  to  either  plane  of  projection,  its 
trace  on  this  plane  may  be  assumed  at  pleasure,  while  its  trace  on 
the  other  plane  must  be  perpendicular  to  AB,  as  in  (2),  Fig.  9  ; 
Tt  is  the  horizontal,  and  Tt'  the  vertical  trace  of  a  plane  perpen- 
dicular to  the  horizontal  plane. 

Also  in  (3),  Tt  is  the  horizontal  and  T*'  the  vertical  trace  of  a 
plane  perpendicular  to  the  vertical  plane,  Art.  (10). 


21.  To  represent  or  assume  a  straight  line,  both  projections 
may  be  drawn  at  pleasure,  as  in  (1),  Fig.  10,  mn  is  the  horizon- 
tal, and  m'n'  the  vertical  projection  of  a  portion  of  a  straight  line 
in  the  first  angle. 

In  (2),  mn  is  the  horizontal  and  m'n'  the  vertical  projection  of 
a  portion  of  a  straight  line  in  the  second  angle.  In  (8),  the  line  is 
represented  in  the  third  angle,  and  in  (4),  in  the  fourth  angle. 

In  Fig.  11  are,  (1),  the  projections  of  a  right  line  parallel  to 
the  horizontal  plane ;  (2),  those  of  a  right  line  parallel  to  the  ver- 
tical plane;   (3),  those  of  a  right  line  perpendicular  to  the  hori* 


DESCRIPTIVE    GEOMETRY.  11 

zontal  plane ;  and  (4),  those  of  a  right  line  perpendicular  to  th 
vertical  plane,  Art.  (14.) 


22.  To  assume  a  point  upon  a  given  right  line;  since  the  pro- 
jections of  the  point  must  be  on  the  projections  of  the  line,  Art.  ' 
(13),  and  in  the  same  perpendicular  to  the  ground  line,  Art.  (19); 
we  assume  the  horizontal  projection  as  m,  Fig.  17,  on  mn,  and 
at  this  point  erect  mm'  perpendicular  to  AB :  m' ,  where  it  inter- 
sects m'n',  will  be  the  vertical  projection  of  the  point. 


23.  If  two  lines  intersect,  their  projections  will  intersect;  fol 
the  point  of  intersection  being  on  each  of  the  lines,  its  horizontal 
projection  must  be  on  the  horizontal  projection  of  each  of  the 
lines,  Art.  (13),  and  hence  at  their  intersection.  For  the  same 
reason,  the  vertical  projection  of  the  point  must  be  at  the  intersec- 
tion of  their  vertical  projections.  These  two  points  being  the 
projection  of  the  same  point,  must  be  in  the  same  straight  line 
perpendicular  to  the  ground  line,  Art.  (19).  Hence  if  any  two 
lines  intersect  in  space,  the  right  line  joining  the  points  in  which 
their  projections  intersect,  must  be  perpendicular  to  the  ground  line. 

Therefore,  to  assume  two  right  lines  which  intersect,  we  draw 
at  pleasure  both  projections  of  the  first  line,  and  the  horizon- 
tal projection  of  the  second  line  intersecting  that  of  the  first ; 
through  this'  point  of  intersection  erect  a  perpendicular  to  the 
ground  line  until  it  intersects  the  vertical  projection  of  this  line ; 
through  this  point  draw  at  pleasure  the  vertical  projection  of 
the  second  line.  Thus,  in  Fig.  16,  assume  mn  and  m'n',  also 
mo ;  through  m  erect  mm'  perpendicular  to  AB,  and  through  m 
draw  m'o'  at  pleasure. 

Two  parallel  right  lines  are  assumed  by  drawing  their  pro 
jections  respectively  parallel,  Art.  (16). 


12  DESCRIPTIVE    UKOMKTEY. 


NOTATION    TO    BE    USED    IN    THE    DESCRIPTION    OF    DRAWINGS. 

24.  Points  represented  as  in  Fig.  7,  will  be  described  as 
the  point  (mm')\  the  letter  at  the  horizontal  projection  being 
always  written  and  read  first,  or  simply  as  the  point  M. 

Planes  given  by  their  traces,  as  in  Fig.  8,  will  be  described 
as  the  plane  tTt\  the  middle  letter  being  the  one  at  the  in- 
tersection of  the  two  traces,  and  the  other  letter  of  the  hori- 
zontal trace  being  the  first  in  order.  If  the  traces  are  parallel 
to  the  ground  line,  or  do  not  intersect  it  within  the  limits  of 
the  drawing,  the  same  notation  will  be  used,  the  middle  let- 
ter being  placed  on.  both  traces;  in  the  first  case,  at  the  left- 
hand  extremity,  and  in  the  second  case,  at  the  extremity  near- 
est the  ground  line! 

Lines  given  by  their  projections,  as  in  Fig.  10,  will  be 
described  as  the  line  (mn,  mV),  the  letters  on  the  horizontal 
projection  being  first  in  order;  or  simply  the  line  MN. 

The  planes  ^  of  projection  will  often  be  described  by  the 
capitals  H  and  V ;  H  denoting  the  horizontal,  and  V  the  ver- 
tical plane. 

The  ground  line  will  be  in  general  denoted  by  AB. 


MANNER  OF  DELINEATING  THE  DIFFERENT  LINES  USED  IN  THB 
REPRESENTATION  OF  MAGNITUDES,  OR  IN  THE  CONSTRUCTION 
OF    PROBLEMS. 

25.  The  projections  of  the  same  point  will  be  connected  by  a 
dotted  line,  thus  . . 

Traces  of  planes  which  are  given  or  required,  when  they  can 
be  seen  from  the  point  of  sight, — that  is,  when  the  view  is  not 
obstructed,  either  by  the  planes  of  projection  or  by  some  inter- 
vening opaque  object, — are  drawn  full.  When  not  seen,  of 
when  they   are  the  traces  of  auxiliary   planes,  not  the  project- 


DESCRIPTIVE   GEOMETRY.  13 

ing  planes  of  right  lines,  they  will  be  drawn  broken  and  dotted, 
thus: 


Lines,  or -portions  of  lines,  either  given  or  required,  when  seen, 
will  have  their  projections  full.  When  not  seen,  or  auxiliary, 
these  projections  will  be  broken,  thus  : 


In  the  construction  of  problems,  planes  or  surfaces  which 
are  required  will  be  regarded  as  transparent,  not  concealing 
other  parts  previously  drawn. 

All  lines  or  surfaces  are  regarded  as  indefinite  in  extent,  un- 
less limited  by  their  form,  or  a  definite  portion  is  considered 
for  a  special  purpose.  Thus  the  ground  line  and  projections 
of  lines  in  Fig.  10  are  supposed  to  be  produced  indefinitely, 
the  lines  delineated  simply  indicating  the  directions. 


CONSTRUCTION  OF  ELEMENTARY  PROBLEMS    RELATING    TO    THE  RIGHT 
LINE    AND    PLANE. 

26.  Having  explained  the  manner  of  representing  with  ac- 
curacy, points,  planes,  and  right  lines,  we  are  now  prepared  to 
represent  the  solution  of  a  number  of  important  problems,  re- 
lating to  these  magnitudes  in  space. 

In  every  problem  certain  points  and  magnitudes  are  given, 
from  which  certain  other  points  or  magnitudes  are  to  be  con- 
structed. 

Let  a  right  line  be  first  drawn  on  the  paper  or  slate  to  rep 
resent  the  ground  line;  then  assume,  as  in  Art.  (19),  &c,  the 


14  DESCRIPTIVE   GEOMETRY. 

representatives  of  the  given  objects.  The  proper  solution  of  the 
problem  will  now  consist  of  two  distinct  parts.  The  first  is  a 
clear  statement  of  the  principles  and  reasoning  to  be  employed 
in  the  construction  of  the  drawing.  This  is  the  analysis  of  the 
problem.  The  second  is  the  construction,  in  proper  order,  of 
the  different  lines  which  are  used  and  required  in  the  problem. 
This  is  the  construction  of  the  problem. 


2 "7.  Problem  1.  To  find  the  points  in  which  a  given  right 
line  pierces  the  planes  of  projection. 

Let  AB,  Fig.  12,  be  the  ground  line,  and  (m»,  mV),  or  simply 
MN,  the  given  line. 

First.  To  find  the  point  in  which  this  line  pierces  the  hori- 
zontal plane. 

Analysis.  Since  the  required  point  is  in  the  horizontal  plane, 
its  vertical  projection  is  in  the  ground  line,  Art.  (8) ;  and  since 
the  point  is  in  the  given  line,  its  vertical  projection'  will  be  in 
the  vertical  projection  of  this  line,  Art.  (13);  hence  it  must  be 
at  the  intersection  of  this  vertical  projection  with  the  ground 
line.  The  horizontal  projection  of  the  required  point  must  be 
in  a  straight  line  drawn  through  its  vertical  projection,  perpen- 
dicular to  the  ground  line,  Art.  (19),  and  also  in  the  horizontal 
projection  of  the  given  line ;  hence  it  will  be  at  the  intersection 
of  these  two  lines.  But  the  point  being  in  the  horizontal  plane, 
is  the  same  as  its  horizontal  projection,  Art.  (8)  ;  hence  the  rule : 
Produce  the  vertical  projection  of  the  line  until  it  intersects  the 
ground  line  ;  at  the  point  of  intersection  erect  a  perpendicular  to 
the  ground  line,  and  produce  it  until  it  intersects  the  horizontal 
projection  of  the  line;  this  point  of  intersection  is  the  required 
point. 

Construction.  Produce  m'n'  to  m' ;  at  m'  erect  the  perpen- 
dicular wi'm,  and  produce  it  to  ?n.     This  is  the  required  point. 

Second.  In  the  above  analysis,  by  changing  the  word  "  verti 
cal"  into  "  horizontal,"  and  the  reverse,  we  have  the  analysis  and 


DESCRIPTIVE   GEOMETRY.  15 

rule  for  finding  the  point  in  which  the  given  line  pierces  the  ver- 
tical plane. 

Construction.  Produce  mnto  o;  at  o  erect  the  perpendicular 
oo\  and  produce  it  to  o'.     This  is  the  required  point. 


28.  Problem  2.  To  find  the  length  of  a  right  line  joining  two 
given  points  in  space. 

Let  AB,  Fig.  13,  be  the  ground  line,  and  {mm')  and  (nnf)  the 
two  given  points. 

Analysis.  Since  the  required  line  contains  the  two  points,  its 
projection  must  contain  the  projections  of  the  points,  Art.  (13). 
Hence,  if  we  join  the  horizontal  projections  of  the  points  by  a 
right  line,  it  will  be  the  horizontal  projection  of  the  line ;  and  if 
we  join  the  vertical  projections  of  the 'points,  we  shall  have  its 
vertical  projection. 

If  we  now  revolve  the  horizontal  projecting  plane  of  the  line 
about  its  horizontal  trace  until  it  coincides  with  the  horizontal 
plane,  and  find  the  revolved  position  of  the  points,  and  join  them 
by  a  right  line,  it  will  be  the  required  distance,  since  the  points 
do  not  change  their  relative  position  during  the  revolution, 
Art.  (17). 

Construction.  Draw  mn  and  m'n'.  MN  will  be  the  required 
line. 

Now  revolve  its  horizontal  projecting  plane  about  mn  until  it 
coincides  with  H;  the  points  M  and  N  will  fall  at  m"  and  ri\  at 
distances  from  m  and  n  equal  to  rm'  and  sn'  respectively,  Art. 
(17) ;  join  m"  and  n" ;   m"n"  will  be  the  required  distance. 

Since  the  point  o,  in  which  the  line  produced  pierces  H,  is  in 
the  axis,  it  remains  fixed.  The  line  m"n"  produced  must  then 
pass  through  o,  and  the  accuracy  of  the  drawing  may  thus  bo 
verified. 

29.  Second  method  for  the  same  problem. 

Analysis.  If  we  revolve  the  horizontal  projecting  plane  of  the 


16  DESCRIPTIVE    GEOMETRY. 

line  about  the  projecting  perpendicular  of  either  of  its  points  until 
it  becomes  parallel  to  the  vertical  plane,  the  line  will,  in  its  re- 
volved position,  be  projected  on  this  pl-ane  in  its  true  length,  Art. 
(14).  If  we  then  construct  this  vertical  projection,  it  will  be  the 
required  distance. 

Construction.  Revolve  the  projecting  plane  about  the  perpen- 
dicular at  m.  The  point  n  describes  the  arc  nl,  until  it  comes  into 
the  line  ml  parallel  to  AB ;  I  will  be  the  horizontal  projection  01 
N  in  its  revolved  position.  Its  vertical  projection  must  be  in  IV 
perpendicular  to  AB ;  and  since,  during  the  revolution,  the  point 
N  remains  at  the  same  distance  above  H,  its  vertical  projection 
must  also  be  in  the  line  n'V  parallel  to  AB,  Art.  (7),  therefore  it 
will  be  at  /'. 

The  point  M  being  in  the  axis  remains  fixed,  and  its  vertical 
projection  remains  at  m!  \  m'V  is  then  the  vertical  projection  of 
MN  in  its  revolved  position,  and  the  true  distance. 

By  examining  the  drawing,  it  will  be  seen  that  the  true  dis- 
tance is  the  hypothenuse  of  a  right-angled  triangle  whose  base  is 
the  horizontal  projection  of  the  line,  and  altitude  the  difference 
between  the  distances  of  its  two  extremities  from  the  horizontal 
plane.  Also,  that  the  angle  at  the  base  is  equal  to  the  angle 
made  by  the  line  with  its  projection,  or  the  angle  made  by  the 
line  with  the  horizontal  plane.  Also,  that  the  length  of  the  line 
is  always  greater  than  that  of  its  projection,  unless  it  is  parallel  to 
the  plane  of  projection. 


30.  Every  right  line  of  a  plane  must  pierce  any  other  plane, 
to  which  it  is  not  parallel,  in  the  common  intersection  of  the 
two ;  hence  every  right  line  of  a  plane,  not  parallel  to  the  hori- 
zontal plane  of  projection,  will  pierce  it  in  the  horizontal  trace  of 
the  plane ;  and  if  not  parallel  to  the  vertical  plane,  will  pierce  it  ir\ 
the  vertical  trace :  hence,  to  assume  a  straight  line  in  a  given 
vlane,  take  a  point  in  each  trace,  and  join  the  two  by  a  right  line ; 
or  otherwise,  draw  the  horizontal  projection  at  pleasure;  at  the 


DESCRIPTIVE   GEOMETRY.  17 

points  where  it  intersects  the  ground  .line  and  the  horizontal  trace 
erect  perpendiculars  to  the  ground  line ;  join  the  point  where  the 
first  intersects  the  vertical  trace  with  the  point  where  the  second 
intersects  the  ground  line :  this  will  be  the  vertical  projection  of 
he  line. 

Thus,  in  Fig.  14,  draw  mn,  also  mm'  and  nn  \  join  mV;   it  will 
be  the  required  vertical  projection. 


31.  Problem  3.      To  pass  a  plane  through  three  given  points. 

Let  M,  N,  and  P,  Fig.  15,  be  the  three  points. 

Analysis.  If  we  join  either  two  of  the  points  by  a  right  line, 
it  will  lie  in  the  required  plane,  and  pierce  the  planes  of  projec- 
tion in  the  traces  of  this  plane,  Art.  (30).  If  we  join  one  of  these 
points  with  the  third  point;  we  shall  have  a  second  line  of  the 
plane.  If  we  find  the  points  in  which  these  lines  pierce  the 
planes  of  projection,  we  shall  have  two  points  of  each  trace.  The 
traces,  and  therefore  the  plane,  will  be  fully  determined. 

Construction.  Join  m  and  n  by  the  straight  line  mn\  also  m' 
and  n'  by  mV.  MN  will  be  the  line  joining  the  first  two  points. 
This  pierces  H  at  A,  and  V  at  v,  as  in  Problem  1.  Draw  also 
np  and  n'p' ;  NP  will  be  the  second  line.  It  pierces  H  at  t  and 
V  at  t\  Join  h  and  t,  by  the  straight  line  ht ;  it  is  the  required 
horizontal  trace.  Join  v  and  t' ;  t'v  is  the  vertical  trace :  Or  pro- 
duce ht  until  it  meets  AB,  and  join  this  point  with  either  v  or 
t'  for  the  vertical  trace,  Art.  (9). 

If  either  MN  or  NP  should  be  parallel  to  AB,  the  plane,  and 
consequently  its  traces,  will  be  parallel  to  AB,  Art.  (9),  and  it 
will  be  necessary  to  find  only  one  point  in  each  trace. 


32.  If  it  be  required  to  pass  a  plane  through  two  right  lines 
which  either  intersect  or  are  .parallel,  we  have  simply  to  find  .the 
points  in  which  these  lines  pierce  the  planes  of  projection,  as  in 
the  preceding  problem.     If  the  lines  do  not  pierce  the  planes  o* 


18  DESCRIPTIVE   GEOMETRY. 

projection,  within  the  limits  of  the  drawing,  then  any  two  points 
of  the  lines  may  be  joined  by  a  right  line,  and  a  point  in  each 
trace  be  determined,  by  finding  the  points  in  which  this  line 
pierces  the  planes  of  projection. 


33.  A  plane  may  be  passed  through  a  point  and  right  line,  by 
joining  the  point  with  any  point  of  the  line  by  a  right  line,  and 
then  passing  a  plane  through  these  lines ;  or  by  drawing  through 
the  point  a  line  parallel  to  the  given  line,  and  then  passing  a 
plane  through  the  parallels,  as  above. 


34.  Problem  4,  To  find  the  angle  between  two  right  lines 
'which  intersect. 

Let  MN  and  MO,  Fig.  16,  be  the  two  right  lines,  assumed  as 
in  Art.  (23). 

Analysis.  Since  the  lines  intersect,  pass  a  plane  through  them, 
and  revolve  this  plane  about  its  horizontal  trace  until  it  coincides 
with  the  horizontal  plane,  and  find  the  revolved  position  of  the 
two  lines.  Since  they  do  not  change  their  relative  position,  their 
angle,  in  this  new  position,  will  be  the  required  angle. 

Construction*  The  line  MN  pierces  H  at  n,  and  the  line  MO 
at  o,  Art.  (27) ;  no  is  then  the  horizontal  trace  of  the  plane  con- 
taining the  two  lines,  Art.  (32).  Revolve  this  plane  about  no 
until  it  coincides  with  H.  The  point  M  falls  at  p,  Art.  (17). 
The  points  n  and  o,  being  in  the  axis,  remain  fixed ;  np  will  then 
be  the  revolved  position  of  MN,  and  po  of  NO,  and  the  angle  npo 
will  be  the  required  angle. 


35.  Second  method  for  the  same  problem. 

Analysis.  Revolve  the  plane  of  the  two  lines  about  its  hori- 
ontal  trace  until  it  becomes  perpendicular  to  the  horizontal 
plane ;    then  revolve  it  about  its  new  vertical  trace  until  it  co- 


DESCRIPTIVE   GEOMETRY,  19 

incides  with  the  vertical  plane ;  the  angle  will  then  be  in  the  ver- 
tical plane  in  its  true  size. 

Construction.  First  revolve  the  plane  about  no  until  it  be- 
comes perpendicular  to  H.  T7'  will  be  the  new  vertical  trace, 
the  point  M  will  be  horizontally  projected  at  s,  and  vertically  at 
s\  s'r  being  equal  to  sp. 

Now  revolve  the  plane  about  Tt'  until  it  coincides  with  V;  o 
will  revolve  to  o",  s  to  s",  and  n  to  n",  while  the  point  M,  or 
(.sV),  will  be  found  at  w  ;  ri'w  and  o"w  will  be  the  revolved  po- 
sitions of  the  two  lines,  and  n"wo"  the  required  angle. 

By  examining  the  drawing,  it  will  be  seen  that  if  the  angle  is 
oblique,  it  is  less  than  its  projection,  unless  both  lines  are  parallel 
to  the  plane  of  projection,  in  which  case  the  angle  is  equal  to  its 
projection. 

Let  the  problem  be  constructed  with  one  of  the  lines  parallel 
to  the  ground  line. 


30.  If  two  right  lines  be  perpendicular  to  each  other  in  space, 
and  one  of  them  parallel  to  the  plane  of  projection,  their  projec- 
tions ivill  be  perpendicular.  For  the  projecting  plane  of  the  line 
which  is  not  parallel  to  the  plane  of  projection  is  perpendicular  to 
the  second  line,  and  also  to  its  projection,  since  this  projection  is 
parallel  to  the  line  itself,  Art.. (14) ;  and  since  this  projection  is 
perpendicular  to  this  projecting  plane,  it  is  perpendicular  to  its 
trace,  which  is  the  projection  of  the  first  line. 


37.  Problem  o.  To  find  the  position  of  a  line  bisecting  the 
angle  formed  by  two  right  lines,  one  of  which  is  perpendicular  to 
either  plane  of  projection. 

Let  MN  and  OP,  Fig.  17,  be  the  two  lines,  the  latcer  being 
perpendicular  to  the  vertical  plane. 

Analysis.  If  the  plane  of  the  two  lines  be  revolved  about  the 
second,  until  it  becomes  parallel  to  the  horizontal  plane,  the  angle 


20  DE  CRIPTCVE   G   OMETltY. 

will  be  projected  on  this  plane  in  its  true  size,  and  may  be  bi- 
sected by  a  right  line.  If  the  plane  be  then  revolved  to  its  primi- 
tive position,  and  the  true  position  of  one  point  of  the  bisecting 
line  be  determined,  and  joined  with  the  vertex  of  the  given  angle, 
we  shall  have  the  required  line. 

Construction.  Let  the  plane  of  the  two  lines  be  revolved  about 
OP,- until  it  becomes  parallel  to  II.  Any  point  of  MN,  as  M,  will 
describe  the  arc  of  a  circle  parallel  to  V,  and  be  horizontally  pro 
jected  at  m'\  and  om"  will  be  the  projection  of  MN,  and  m"op 
will  be  the  true  size  of  the  angle.  Bisect  it  by  oq,  which  will  be 
the  horizontal  projection  of  the  bisecting  line  in  its  revolved  posi- 
tion. Join  in"  with  any  point  of  o/>,  as  p ;  this  will  be  the  hon 
zontal  projection  of  a  line  of  the  given  plane,  in  its  revolved 
position,  which  intersects  the  bisecting  line  in  a  point  horizontally 
projected  at  q.  When  the  plane  resumes  its  primitive  position, 
this  line  will  be  horizontally  projected  in  mp.  and  the  point,  of 
which  q  is  the  horizontal  projection,  will  be  horizontally  projected 
at  r,  and  vertically  at  r  ;  hence  or  will  be  the  horizontal,  and 
om'  the  vertical  projection  of  the  required  line. 

Or  the  plane  of  the  two  lines  may  be  revolved  about  its  verti- 
cal trace,  and  the  true  position  determined1  as  indicated  in  the 
figure. 

38.  Problem  6.     To  find  the  intersection  of  two  planes. 

Let  tTt'  and  sS/,  Fig.  18,  be  the  two  planes. 

Analysis.  Since  the  line  of  intersection  is  a  right  line,  con- 
tained in  each  plane,  it  must  pierce  the  horizontal  plane  in  the 
horizontal  trace  of  each  plane,  Art.  (30)  ;  that  is,  at  the  inter 
section  of  the  two  traces.  V  r  the  same  reason,  it  must  pierce 
the  vertical  plane  at  the  irt* /section  of  the  vertical  traces.  Ji 
these  two  points  be  joined  by  a  right  line,  it  will  be  the  required 
intersection. 

Construction.  The  required  line  pierces  H  at  o  and  V  at  p' 
o  is  its  own  horizontal  projection,  and  p'  is  horizontally  pro 
jected   at  p ;  hence  po  is  the  horizontal    projection   of  the    re 


DfcSCRIPTIVK    GKOMKTRY.  21 

quired  line ;  o  is  vertically  projected  at  o' ;  p'  is  its  own  verti- 
cal projection ;  and  o'p'  is  the  vertical  projection  of  the  required 
line. 


39.  Second  Method  for  the  same  problem.  When  either 
the  horizonta.  or  vertical  traces  do  not  intersect  within  the 
limits  of  the  drawing.  Let  fit'  and  *S#',  Fig.  19,  be  the 
planes;  tT  and  sS  not  intersecting  within  the  limits  of  the 
drawing. 

Analysis.  If  we  pass  any  plane  parallel  to  the  vertical  plane, 
it  will  intersect  each  of  the  given  planes  in  a  line  parallel  to  its 
vertical  trace,  and  these  two  lines  will  intersect  in  a  point  of  the 
required  intersection.  A  second  point  may  be  determined  in 
the  same  way,  and  the  right  line  joining  these  two  points  will 
be  the  required  line. 

Construction.  Draw  pq  parallel  to  AB ;  it  will  be  the  trace 
of  an  auxiliary  plane.  It  intersects  the  two  given  planes  in 
lines,  which  pierce  H  at  p  and  q,  and  are  vertically  projected 
in  p'o'  and  q'o' ;  o'  is  the  vertical,  and  o  the  horizontal  pro- 
jection of  their  intersection.  Draw  mn  also  parallel  to  AB,  and 
thus  determine  L.  OL  is  the  required  line.  Let  the  problem  be 
constructed  when  both  planes  are  parallel  to  the  ground  line. 

40.  Problem  7.  To  find  the  point  in  which  a  given  right  line 
pierces  a  given  plane. 

Let  MN,  Fig.  20,  be  the  given  line,  and  tTt'  the  given  plane. 

Analysis.  If  through  the  line  any  plane  be  passed,  it  will 
intersect  the  given  plane  in  a  right  line,  which  must  contain  the 
required  point,  Art.  (30)  This  point  must  also  be  on  the 
given  line  ;  hence  it  will  be  at  the  intersection  of  the  two  lines. 

Construction.  Let  the  auxiliary  plane  be  the  horizontal 
projectisg  plane  of  the  line ;  np  is  its  horizontal  and  pt'  its 
vertical  trace,  Art.  (10).  It  intersects  tTt'  in  a  right  line,  which 
pierces  II  at  o  and  V  at  t\  of  which  o't'  is  the  vertical  projec- 


22  DESCRIPTIVE   GEOMETRY. 

tion,  Art.  (38).  The  point  ?n',  in  which  o't'  intersects  mV,  is 
the  vertical  projection  of  the  required  point;  and  m  is  its  hori- 
zontal projection.  The  accuracy  of  the  drawing  may  be  verified 
by  using  the  vertical  projecting  plane  of  MN,  as  an  auxiliary 
plane,  and  determining  m  directly  as  represented  in  the  figure. 

Let  the  problem  be  constructed  when  the  given  line  is  parallel 
to  the  ground  line. 

41.  Second  Method  for  the  same  problem.  When  the  plane 
is  given  by  any  two  of  its  right  lines,  find  the  points  in  which 
these  two  lines  pierce  either  projecting  plane  of  the  given  line, 
and  join  these  points  by  a  straight  line ;  this  will  intersect  the 
given  line  in  the  required  point. 

Construction.  Let  MN  and  OP,  Fig.  21,  be  the  lines  of  the 
given  plane,, intersecting  at  L,  and  QR  be  the  given  line.  The 
line  MN  pierces  the  horizontal  projecting  plane  of  QR  at  a 
point  of  which  m  is  the  horizontal,  and  m'  the  vertical  projec- 
tion. OP  pierces  the  same  plane  at  P,  and  p'm'  is  the  vertical 
projection  of  the  line  joining  these  two  points.  This  intersects 
q'r'  at  r',  which  is  the  vertical  projection  of  the  required  point, 
and  r  its  horizontal  projection. 

42.  If  either  projection  of  a  point  of  an  oblique  plane  be 
given,  the  other  projection  may  at  once  be  determined  by  a 
simple  application  of  the  principles  of  the  preceding  problem. 
Thus  let  m,  Fig.  22,  be  the  horizontal  projection  of  a  point  of 
the  plane  tTt'.  If  at  m  a  perpendicular  be  erected  to  H,  it  will 
pierce  f£t'  in  the  only  point  of  the  plane  which  can  be  hori- 
zontally projected  atm;  m  is  the  horizontal,  and  m"m'  the  ver 
tical  projection  of  this  perpendicular.  Through  it  pass  any 
plane,  as  that  whose  horizontal  trace  is  no.  Since  this  plane  is 
perpendicular  to  H,  nn'  will  be  its  vertical  trace.  It  intersects 
fit'  in  a  right  line,  of  which  o'n'  is  the  vertical  projection ;  hence 
mf  is  the  required  vertical  projection,  Art.  (40.) 


DESCRIPTIVE   GEOMETRY.  23 

The  auxiliary  plane  may  be  passed  parallel  to  f£ ;  mp  will  be 
its  horizontal,  and  pp'  its  vertical  trace.  It  intersects  f£t'  in  a 
line  parallel  to  IT,  which  pierces  V  at  p',  of  which  mp  is  the 
horizontal,  and  m'p'  the  vertical  projection,  and  m'  will  be  the 
required  vertical  projection. 

In  a  similar  way,  if  the  vertical  projection  be  given,  the  hori- 
Eontal  can  be  found. 


43.  If  a  right  line  is  perpendicular  to  a  plane,  its  projections 
will  be  respectively  perpendicular  to  the  traces  of  the  plane. 

For  the  horizontal  projecting  plane  of  the  line  is  perpendicu- 
lar to  the  given  plane,  since  it  contains  a  line  perpendicular  to 
it.  This  projecting  plane  is  also  perpendicular  to  the  horizon- 
tal plane,  Art.  (11).  It  is  therefore  perpendicular  to  the  in- 
tersection of  these  two  planes,  which  is  the  horizontal  trace  of 
the  given  plane.  Hence  the  horizontal  projection  of  the  line, 
which  is  a  line  of  this  projecting  plane,  must  be  perpendicular 
to  the  horizontal  trace. 

In  the  same  way  it  may  be  proved  that  the  vertical  projection 
of  the  line  will  be  perpendicular  to  the  vertical  trace. 

Conversely,  if  the  projections  of  a  right  line  are  respectively 
perpendicular  to  the  traces  of  a  plane,  the  line  will  be  perpendicular 
to  the  plane. 

For,  if  through  the  horizontal  projection  of  the  line,  its  hori- 
zontal projecting  plane  be  passed,  it  will  be  perpendicular  to  the 
horizontal  trace  of  the  given  plane,  and  therefore  perpendicular 
to  the  plane.  In  the  same  way  it  may  be  proved  that  the  ver- 
tical projecting  plane  of  the  line  is  perpendicular  to  the  given 
plane;  therefore  the  intersection  of  these  two  planes,  which  is 
the  given  line,  is  perpendicular  to  the  given  plane. 

Hence,  to  assume  a  right  line  perpendicular  to  a  plane,  we 
draw  its  projections  perpendicular  to  the  traces  of  the  plane 
respectively. 

Also,  to  assume  a  plane  perpendicular  to  n  right  line,  we  draw 


24  DESCKIPTIVE   GEOMETRY. 

the  two  traces  from  any  point  in  the  ground  line,  perpendicular 
to  the  projections  of  the  line. 


44.  Problem  8.  To  draw  through  a  given  point  a  right  line 
perpendicular  to  a  given  plane7  and  to  find  the  distance  of  the 
point  from  the  plane. 

Let  M,  Fig.  23,  be  the  given  point,  and  tTtr  the  plane. 

Analysis.  Since  the  required  perpendicular  is  to  pass  through 
the  given  point,  its  projections  must  pass  through  the  projec- 
tions of  the  point,  Art.  (13);  and  since  it  is  to  be  perpendicular 
to  the  plane,  these  projections  must  be  respectively  perpendic- 
ular to  the.  traces  of  the  plane,  Art.  (43).  Hence,  if  through 
the  horizontal  projection  of  the  point,  a  right  line  be  drawn 
perpendicular  to  the  horizontal  trace,  and  through  the  vertical 
projection,  a  right  line  perpendicular  to  the  vertical  trace,  they 
will  be  respectively  the  horizontal  and  vertical  projections  of 
the  required  line. 

If  the  point  in  which  this  perpendicular  pierces  the  plane  be 
found,  the  distance  between  this  point  and  the  given  point  will 
be  the  required  distance  or  length  of  the  perpendicular. 

Construction.  Through  m  draw  mn  perpendicular  to  tfT,  and 
through  mfr  mfn',  perpendicular  to  trT.  MN  will  be  the  re- 
quired perpendicular.  N  is  the  point  in  which  MN  pierces 
the  plane,  Art.  (40),  and  m"nn  the  length  of  the  perpendicular, 
Art.  (28). 

Let  the  problem  be  constructed  when  the  plane  is  parallel  to 
the  ground  line ;  also  when  it  is  perpendicular  to  it. 


45.  Problem  9.  To  p-qject  a  given  right  line  on  any  oblique 
2)lane,  and  to  show  the  true  position  of  this  projection. 

Let  MN,  Fig.  24,  be  the  given  line,  and  tTt'  the  given  plane. 

Analysis.  If  through  any  two  points  of  the  line  perpendic- 
ulars  be   drawn    to   the   plane,    and    the  points  in   which    they 


DESCRIPTIVE    GEOMETRY.  25 

pierce  the  plane  be  found,  these  will  be  two  points  of  the  re- 
quired projection,  and  the  right  line  joining  them  will  be  the 
required  line.  If  now  the  plane  be  revolved  about  its  horizon- 
tal trace  to  coincide  with  the  horizontal  plane,  or  about  its  ver- 
tical trace  until  it  coincides  with  the  vertical  plane,  and  the 
revolved  position  of  these  two  points  be  found  and  joined  by  a 
right  line,  this  will  show  the  true  position  of  the  line  in  the 
oblique  plane. 

Construction.  Assume  the  two  points  M  and  P,  Art.  (22), 
and  draw  the  perpendiculars  MR  and  PS,  Art.  (44).  The  first 
pierces  the  plane  at  R,  and  the  second  at  S,  Art.  (40) ;  and  rs 
will  be  the  horizontal,  and  rV  the  vertical  projection  of  the  re- 
quired projection.  The  point  N,  in  which  the  given  line  pierces 
the  plane,  will  also  be  one  point  of  the  required  projection. 

Now  revolve  the  plane  about  tT  until  it  coincides  with  H. 
R  is  found  at  r",  Art  (17),  and  S  at  s'\  and  r"s"  is  the  true 
position  of  RS  in  its  own  plane :  r"s"  produced  must  pass  through 
the  point  in  which  the  projection  pierces  H. 

If  the  given  line  be  parallel  to  the  plane,  it  will  only  be  neces- 
sary to  determine  the  projection  of  one  point  on  the  plane,  and 
through  this  to  draw  a  line  parallel  to  the  given  line,  Art.  (14). 


46.    Problem    10.       Through  a  given  point,   to  pass  a  plane 
perpendicular  to  a  given  right  line. 

Let  M,  Fig.  25,  be  the  given  point,  and  NO  the  given  line. 

Analysis.  Since  the  plane  is  to  be  perpendicular  to- the  line, 
its  traces  must  be  respectively  perpendicular  to  the  projections  of 
the  line,  Art.  (43).  We  thus  know  the  direction  of  the  traces. 
Through  the  point,  draw  a  line  parallel  to  the  horizontal  trace ; 
it  will  be  a  line  of  the  required  plane,  and  will  pierce  the  vertical 
plane  in  a  point  of  the  vertical  trace.  Through  this  point  draw  a 
right  line  perpendicular  to  the  vertical  projection  of  the  line ;  it 
will  be  the  vertical  trace  of  the  required  plane.  Through  th 
point  in  which  this  trace  intersects  the  ground  line,  draw  a  righ 


26  DESCRIPTIVE    GEOMETRY. 

line  perpendicular  to  the  horizontal  projection  of  the  line ;  it  will 
be  the  horizontal  trace. 

Construction.  Through  m  draw  mp}  perpendicular  to  no ;  it 
will  be  the  horizontal  projection  of  a  line  through  M,  parallel  to 
the  horizontal  trace  ;  and  since  this  line  is  parallel  to  II,  its  ver- 
tical projection  will  be  m'p\  parallel  to  AB.  This  line  pierces  V 
at  p;,  Art.  (27).  Draw  pfT  perpendicular  to  n'o\  and  Tt  perpen- 
dicular to  no ;  tTp'  will  be  the  required  plane.  Or,  through  M, 
draw  MS  parallel  to  the  vertical  trace.  It^pierces  H  at  s,  which 
must  be  a  point  of  the  horizontal  trace,  and  the  accuracy  of  the 
drawing  may  thus  be  tested. 

47.  Pboblem  11.  To  pass  a  plane  through  a  given  point, 
parallel  to  two  given  right  lines. 

Let  M,  Fig.  26,  be  the  point,  and  NO,  and  PQ,  the  two  given 
lines. 

Analysis.  Through  the  given  point,  draw  a  line  parallel  to 
each  of  the  given  lines.  The  plane  of  these  two  lines  will  be  the 
required  plane,  since  it  contains  a  line  parallel  to  each  of  the 
given  lines. 

Construction.  Through  m  draw  ms,  parallel  to  no,  and 
through  m',  mV  parallel  to  n'o  .  The  line  MS  will  be  parallel 
to  NO,  Art.  (16).  In  the  same  way,  construct  MR  parallel  to 
QP.  These  lines  pierce  H  at  5  and  t  respectively,  and  MR 
pierces  V  at  r' ;   hence  tTr'  is  the  required  plane,  Art.  (82). 

Let  the  problem  be  constructed  when  one  of  the  given  lines  is 
parallel  to  the  ground  line. 

Let  the  problem,  to  pass  a  plane  through  a  given  point  parallel 
to  a  given  plane,  also  be  constructed. 

48.  Problem  12.  To  pass  a  plane  through  a  given  right  line, 
parallel  to  another  right  line. 

Let  MN,  Fig.  27,  be  the  line  through  which  the  plane  is  to  be 
passed,  and  PQ  the  other  given  line. 


DESCRIPTIVE   GEOMETRY.  27 

Analysis.  Through  any  point  of  the  first  line,  draw  a  line 
parallel  to  the  second.  Through  this  auxiliary  line  and  the  first, 
pass  a  plane.     It  will  be  the  required  plane. 

Construction.  Through  R,  on  the  first  line,  draw  RO  parallel 
to  PQ.  It  pierces  H  at  o,  and  V  at  tf.  MN  pierces  II  at  m,  and 
V  at  n' ;    hence  oTt'  is  the  required  plane. 

Let  the  problem  be  constructed  when  either  line  is  parallel  to 
the  ground  line. 


49.  Problem  13.  To  find  the  shortest  distance  from  a  given 
point  to  a  given  right  line. 

Let  M,  Fig.  28,  be  the  given  point,  and  NO  the  given  straight 
line. 

Analysis.  The  required  distance  is  the  length  of  a  perpendic- 
ular from  the  point  to  the  line.  If  through  the  given  point  and 
the  line  we  pass  a  plane,  and  revolve  this  plane  about  either 
trace  until  it  coincides  with  the  corresponding  plane  of  projection, 
the  line  and  point  will  not  change  their  relative  positions ;  hence, 
if  through  the  revolved  position  of  the  point  we  draw  a  perpen- 
dicular to  the  revolved  position  of  the  line,  it  will  be  the  required 
distance. 

Construction.  Through  M  draw  MP  parallel  to  NO.  It 
pierces  H  at  p.  NO  pierces  H  at  o.  po  is  then  the  horizontal 
trace  of  the  plane  through  M  and  NO,  Art.  (32).  Revolve  this 
plane  about  op  until  it  coincides  with  H.  M  tails  at  nt",  Art. 
(17).  Since  p  remains  fixed,  pm"  is  the  revolved  position  of  MP. 
NO  being  parallel  to  MP  before  revolution,  will  be  parallel  after; 
and  as  o  is  in  the  axis,  oq"  parallel  to  pm"  will  be  the  revolved 
position  of  NO.  Draw  m"q"  perpendicular  to  oq"\  it  will  be  the 
required  distance.  When  the  plane  is  revolved  back  to  its  primi 
tive  position,  m"  is  horizontally  projected  at  m,  and  q"  at  q 
hence  MQ  is  the  perpendicular  in  its  true  position. 


28  P.K8CRIPTIVK    GKOMKTIiY. 

50.  Second  method  for  the  same  problem. 

Analysis.  If  through  the  given  point  a  plane  be  passed  per- 
pendicular to  the  given  line,  Art.  (46),  and  the  point  in  which 
the  given  line  pierces  the  plane  be  found,  Art.  (40),  and  joined 
with  the  given  point,  we  shall  have  the  required  distance,  the 
true  length  of  which   can  be  found  as  in  Art.  (28). 

Let  the  problem  be  constructed  in  accordance  with,  this 
analysis.  . 

Let  the  problem  also  be  constructed  when  the  given  line  is 
parallel  to  the  horizontal   plane. 


51.  Problem  14.  To  find  the  angle  which  a  given  right  line 
makes  with  a  given  plane. 

Let  MN,  Fig.  29,  be  the  given  line,  and  tTt'  the  given  plane. 

Analysis.  The  angle  made  by  the  line  with  the  plane,  is  the 
same  as  that  made  by  the  line  with  its  projection  on  the  plane. 
Hence,  if  through  any  point  of  the  line  a  perpendicular  be  drawn 
to  the  plane,  the  foot  of  this  perpendicular  will  be  one  point  of 
the  projection.  If  this  point  be  joined  with  the  point  in  which 
the  given  line  pierces  the  plane,  we  shall  have  the  projection  of 
the  line  on  the  plane,  Art.  (45).  This  projection,  the  perpen- 
dicular, and  a  portion  of  the  given  line,  form  a  right-angled 
triangle;  of  which  the  projection  is  the  base,  and  the  angle  at 
the  base  is  the  required  angle.  But  the  angle  at  the  vertex,  that 
is,  the  angle  between  the  perpendicular  and  given  line,  is  the  com- 
plement of  the  required  angle  ;  hence,  if  we  find  the  latter  angle, 
and  subtract  it  from  a  right  angle,  we  shall  have  the  required 
angle. 

Construction.  Through  M  draw  the  perpendicular  MP  to  lTi\ 
Art.  (44).  It  pierces  II  in  p.  The  given  line  pierces  II  in  u, 
and  op  is  the  horizontal  trace  of  the  plane  of  the  two  lines,  Art. 
(32).  Revolve  this  plane  about  op,  and  determine  their  angle; 
pm"o,  as  in  Art.  (34).  Its  complement,  prm' ,  is  equal  to  the 
required  angle. 


DESCRIPTIVE    GEOMETRY.  ZV 

Let  the  problem  be  constructed  when  the  plane  is  parallel  tc 
the  ground  line. 

52.  Problem  15.     To  find  the  angle  between  two  given  planes. 

Let  -s-S/,  Fig.  30,  and  tTt',  be  the  two  planes,  intersecting  in 
the  line  ON,  Art.  (38). 

Analysis.  If  we  pass  a  plane  perpendicular  to  the  intersection 
of  the  two  planes,  it  will  be  perpendicular  to  both ;  and  cut 
from  each  a  right  line  perpendicular  to  this  intersection  at  a 
common  point.  The  angle  between  these  lines  will  be  the 
measure  of  the  required  angle. 

Cofistruction.  Draw  pq  perpendicular  to  on ;  it  will  be  the 
horizontal  trace  of  a  plane  perpendicular  to  ON,  Art.  (43).  This 
plane  intersects  the  given  planes  in  right  lines,  one  of  which 
pierces  H  at  p,  and  the  other  at  q.  If  right  lines  be  drawn 
from  these  points  to  the  point  in  which  the  auxiliary  plane  inter 
sects  ON,  they  will  be  the  lines  cut  from  the  planes,  and  the 
angle  between  them  will  be  the  required  angle. 

The  vertical  trace  of  the  auxiliary  plane  may  be  drawn  as  in 
Art.  (43),  and  the  vertex  of  the  angle  found  as  in  Art.  (40),  and 
then  the  angle  as  in  Art.  (34).  Or  otherwise,  thus  :  Suppose  a 
right  line  to  be  drawn  .from  r  to  the  vertex  of  the  angle,  it  will 
be  perpendicular  to  ON,  since  it  is  contained  in  a  plane  perpen- 
dicular to  it;  it  wili  also  be  perpendicular  to pg,  since  it  is  in  the 
horizontal  projecting  plane  of  ON,  which  is  perpendicular  to 
pq,  Art.  (43).  If  this  projecting  plane  be  revolved  about  no 
until  it  coincides  with  H,  n'  will  fall  at  n"  \  and  since  o  is  fixed, 
on"  will  be  the  revolved  position  of  ON,  and  rm",  perpendicular 
to  on",  will  be  the  revolved  position  of  the  line  joining  r  with 
the  vertex.  If  now  the  plane  of  the  two  lines  be  revolved  about 
pq  until  it  coincides  with  H,  m"  will  be  at  v,  rv  being  equal  to 
rm",  and  pvq  will  be  the  required  angle,  Art.  (34). 

The  point  ?n",  from  its  true  position,  is  horizontally  projected 
at  m,  and  vertically  at  ra',  and  pmq  is  the  horizontal,  and  p'm'q 
the  vertical  projection  of  the  angle. 


30 


DESCRIPTIVE   GEOMETRY. 


Let  the  problem  be  constructed  when  both  planes  are  parallel 
to  the  ground  line. 


53.  If  the  angle  between  a  given  plane  and  either  plane  of 
projection,  as  the  horizontal,  be  required,  we  simply  pass  a  plane 
perpendicular  to  the  horizontal  trace,  as  in  Fig.  31.  This  plane 
cuts  on  from  H,  and  ON  from  fit',  and  the  angle  non",  found 
by  revolving  the  auxiliary  plane  about  on,  Art.  (34),  will  be  the 
required  angle. 

In  the  same  way  the  angle  p'q'p",  between  the  given  plane 
and  vertical  plane,  may  be  found. 


54.  Problem  16.  Either  trace  of  a  plane  being  given,  and 
the  angle  which  the  plane  makes  with  the  corresponding  plane  of 
projection,  to  construct  the  other  trace. 

Let  *T,  Fig.  31,  be  the  horizontal  trace  of  the  plane,  and  def 
the  angle  which  the  plane  makes  with  the  horizontal  plane. 

Analysis.  If  a  right  line  be  drawn  through  any  point  of  the 
given  trace,  perpendicular  to  it,  it  will  be  the  horizontal  trace 
of  a  plane  perpendicular  to  the  given  trace,  and  if  at  the  same 
point  a  line  be  drawn,  making  with  this  line  an  angle  equal  to 
the  given  angle,  this  will  be  the  revolved  position  of  a  line  cut 
from  the  required  plane  by  this  perpendicular  plane,  Art.  (53). 
If  this  line  be  revolved  to  its  true  position,  and  the  point  in 
which  it  pierces  the  vertical  plane  be  found,  this  will  be  a  point 
of  the  required  vertical  trace.  If  this  point  be  joined  with  the 
point  where  the  horizontal  trace  intersects  the  ground  line,  we 
shall  have  the  vertical  trace. 

Construction.  Through  o  draw  no  perpeudicular  to  ^T ;  also 
on" ,  making  the  angle  non"  =z  def;  on"  will  be  the  revolved 
position  of  a  line  of  the  required  plane.  When  this  line  is  re- 
volved to  its  true  position.,  it  pierces  V  at  n\  and  n'T  is  the  re- 
quired trace. 


DESCRIPTIVE   GEOMETRY.  31 

If  the  given  trace  does  not  intersect  the  ground  line  within 
the  limits  of  the  drawing,  the  same  construction  may  be  made 
at  a  second  point  of  the  trace,  and  thus  another  point  of  the 
vertical  trace  be  determined. 


55.  Pkoblem  17.  To  find  the  shortest  line  which  can  be 
drawn,  terminating  in  two  right  lines,  not  in  the  same  plane. 

Let  MN,  Fig.  32,  and  OP,  be  the  two  right  lines. 

Analysis.  The  required  line  is  manifestly  a  right  line,  per- 
pendicular to  both  of  the  given  lines.  If  through  one  of  the 
lines  we  pass  a  plane  parallel  to  the  other,  and  then  project  this 
second  line  on  this  plane,  this  projection  will  be  parallel  to  the 
line  itself,  Art.  (14),  and  therefore  not  parallel  to  the  first  line. 
It  will  then  intersect  the  first  line  in  a  point.  If  at  this  point 
we  erect  a  perpendicular  to  the  plane,  it  will  be  contained  in 
the  projecting  plane  of  the  first  line,  be  perpendicular  to  both 
lines,  and  intersect  them  both.  That  portion  included  between 
them  is  the  required  line. 

Construction.  Through  MN  pass  a  plane  parallel  to  OP, 
Art.  (48)  :  mr  is  its  horizontal,  and  k'ri  its  vertical  trace. 
Through  any  point  of  OP,  as  Q,  draw  QTJ  perpendicular  to 
this  plane,  Art.  (49).  It  pierces  the  plane  at  U,  Art.  (40) ; 
and  this  is  one  point  of  the  projection  of  OP  on  the  parallel 
plane.  Through  U  draw  UX  parallel  to  OP;  it  will  be  the 
projection  of  OP  on  the  plane.  It  intersects  MN  in  X,  which 
is  the  point  through  which  the  required  line  is  to  be  drawn ; 
and  XY,  perpendicular  to  the  plane,  is  the  required  line,  the 
true  length  of  which  is  x"y",  Art.  (28). 

Let  the  problem  be  constructed  with  one  of  the  lines  parallel 
to  the  ground  line. 

Also  with  one  of  the  lines  perpendicular  to  either  plane  oi 
projection. 


32  DESCRIPTIVE    GEOMETRY. 

56.  Second  construction  of  the  same  problem. 

Let  MN  and  OP,  Fig.  33,  be  the  right  lines. 

Through  MN  pass  a  plane  parallel  to  OP,  Art.  (48)  :  mr  is 
its  horizontal  trace.  Through  p  conceive  a  perpendicular  to  be 
drawn  to  this  plane.  The  point  in  which  it  pierces  the  plane 
will  be  one  point  of  the  projection  of  OP  on  the  plane.  To 
find  this  point,  through  the  perpendicular  pass  a  plane  perpen- 
dicular to  OP;  pq  will  be  its  horizontal  trace,  Art.  (43).  This 
plane  will  intersect  the  parallel  plane  in  a  right  line,  which 
pierces  II  at  q.  It  intersects  the  horizontal  projecting  plane  of 
OP  in  a  right  line  perpendicular  to  OP  at  p.  To  determine 
this  line,  revolve  the  projecting  plane  of  OP  about  op  until  it 
coincides  with  II,  Any  point  of  OP,  as  L,  falls  at  /",  and  pi" 
is  the  revolved  position  of  OP.  This  projecting  plane  intersects 
the  parallel  plane  in  a  right  line,  which  pierces  H  at  k,  and  is 
parallel  to  OP ;  ku,  parallel  to  pi" ,  is  the  revolved  position  of 
this  parallel  line  ;  pu,  perpendicular  to  pi",  is  the  revolved  posi- 
tion of  the  intersection  of  the  projecting  plane  and  perpendicular 
plane;  and  u  is  the  revolved  position  of  a  point  of  the  line  of 
intersection  of  the  perpendicular  and  parallel  plane.  Now  re- 
volve the  plane  perpendicular  to  OP  about  pq  as  an  axis,  until  it 
coincides  with  H.  The  point,  of  which  u  is  the  revolved  position, 
falls  at  u",  and  u"q  is  the  revolved  position  of  the  line  of  inter- 
section of  the  perpendicular  and  parallel  plane ;  pp"  is  the  re- 
volved position  of  the  line  through  p  perpendicular  to  the  parallel 
plane,  and  is  equal  to  the  distance  required  ;  and  p"  is  the  re- 
volved position  of  the  projection  of  p  on  the  parallel  plane.  In 
the  counter-revolution,  the'  point  p"  will  be  horizontally  pro- 
jected, somewhere  in  the  perpendicular  to  the  axis  pq ;  p"  x"  is 
the  horizontal  projection  of  the  projection  of  OP  on  the  parallel 
plane,  and  xy,  perpendicular  to  mr,  is  the  horizontal,  and  x'y'  the 
vertical  projection  of  the  required  line. 


DESCRIPTIVE    GEOMETRY.  33 


CONSTRUCTION    AND    CLASSIFICATION    OP   LINES. 

57.  Every  line  may  be  generated  by  the  continued  motion  of  a 
voint.  If  the  generating  point  be  taken  in  any  position  on  the 
line,  and  then  be  moved  to  its  next  position,  these  two  points  may 
be  regarded  as  forming  an  infinitely  small  right  line,  or  elemen- 
tary line.  The  two  points  are  consecutive  points,  or  points  hav- 
ing no  distance  between  them,  and  may  practically  be  considered 
as  one  point. 

The  line  may  thus  be  regarded  as  made  up  of  an  infinite  nunn 
ber  of  infinitely  small  elements,  each  element  indicating  the 
direction  of  the  motion  of  the  point  while  generating  that  part 
of  the  line. 


58.  The  law  which  directs  the  motion  of  the  generating  point, 
determines  the  nature  and  class  of  the  line. 

If  the  point  moves  always  in  the  same  direction,  that  is,  so  that 
the  elements  of  the  line  are  all  in  the  same  direction,  the  line  gen- 
erated is  a  right  line. 

If  the  point  moves  so  as  continually  to  change  its  direction 
from  point  to  point,  the  line  generated  is  a  curved  line  or  curve. 

If  all  the  elements  of  a  curve  are  in  the  same  plane,  the  curve  is 
of  single  curvature. 

If  no  three  consecutive  elements,  that  is,  if  no  four  consecutive 
points  are  in  the  same  plane,  the  curve  is  of  double  curvature. 

We  thus  have  three  general  classes  of  lines. 

I.  Right  lines  :  all  of  whose  points  lie  in  the  same  direction. 

II.  Curves  of  single  curvature  :  all  of  whose  points  lie  in 
the  same  plane. 

III.  Curves  of  double  curvature  :  no  four  consecutive  points 
of  which  lie  in  the  same  plane. 


59.  The  simplest  curves  of  single  curvature  are  : 

a 


34:  DESCRIPTIVE  GEOMETRY* 

I.  The  circumference  of  a  circle,  which  may  be  generated  by  a 
point  moving  in  the  same  plane>  so  as  to  remain  at  the  same  dis* 
tance  from  a  given  point. 

II.  A  parabola,  which  may  be  generated  by  a  point  moving 
in  the  same  plane,  so  that  its  distance  from  a  given  point  shall 
be  constantly  equal  to  its  distance  from  a  given  right  line. 

The  given  point  is  the  focus,  the  given  right  line  the  directrix. 

If  through  the  focus  a  right  line  be  drawn  perpendicular  to 
the  directrix,  it  is  the  axis  of  the  parabola ;  and  the  point  in 
which  the  axis  intersects  the  curve  is  the  vertex* 

From  the  definition,  the  curve  may  readily  be  constructed  by 
points,  thus  :  Let  F,  Fig.  34,  be  the  focus,  and  CD  the  directrix* 
Through  F  draw  FC  perpendicular  to  CD.  It  will  be  the  axis. 
The  point  Y,  midway  between  F  and  C,  is  a  point  of  the  curve, 
and  is  the  vertex.  Take  any  point  on  the  axis,  as  P,  and  erect 
the  perpendicular  PM  to  the  axis.  With  F  as  a  centre,  and  CP 
as  a  radius,  describe  an  arc  cutting  PM  in  the  two  points  M  and 
Mr.     These  will  be  points  of  the  curve>  since 

FM  =*  CP  =*  DM,  also  FM'  =  CP  *=  D'M\ 

In  the  same  way  all  the  points  may  be  constructed* 

III.  An  ellipse,  which  may  be  generated  by  a  point  moving  in 
tne  same  plane,  so  that  the  sum  of  its  distances  from  two  fixed 
points  shall  be  constantly  equal  to  a  given  right  line. 

The  two  fixed  points  are  the  foci.  The  curve  may  be  construct- 
ed by  points,  thus :  Let  F  and  F',  Fig.  35,  be  the  two  foci,  and 
YY'  the  given  right  line,  so  placed  that  YF  =  Y'F'. 

Take  any  point  as  P  between  F  and  F\  With  F  as  a  centre, 
and  Y'P  as  a  radius,  describe  an  arc*  With  F'  as  a  centre,  and 
VP  as  a  radius,  describe  a  second  arc,  intersecting  the  first  in 
the  points  M  and  M'.  These  will  be  points  of  the  required  curve, 
since 


DESCRIPTIVE   GEOMETRY  35 

MF  +  MF'  =  VP  +  V'P  =  YV';  also  MF+  M'F'  =  VY'. 

In  the  same  way  all  the  points  may  be  constructed.  Y  and  V' 
are  evidently  points  of  the  curve,  since 

YF  +  VF'  =  Y'F'  +  YF'  =  W;  also  Y'F'  +  Y'F  =  W. 

The  point  C,  midway  between  the  foci,  is  the  centre  of  the 
curve.  The  line  YY',  passing  through  the  foci,  and  terminating 
in  the  curve,  is  the  transverse  axis  of  the  curve. 

The  points  Y  and  Y'  are  the  vertices  of  the  curve.  DD'  per- 
pendicular to  YY',  at  the  centre,  is  the  conjugate  axis  of  the  curve. 

If  the  two  axes  are  given,  the  foci  may  be  constructed  thus : 
With  D  the  extremity  of  the  conjugate  axis  as  a  centre,  and  CV 
the  semi-transverse  axis  as  a  radius,  describe  an  arc  cutting  YY' 
in  F  and  F'.     These  points  will  be  the  foci,  for 

DF  +  DF'  =  2CY  &  YV'. 

IY.  The  hyperbola,  which  may  be  generated  by  moving  a  point 
in  the  same  plane,  so  that  the  difference  of  its  distances  from  two 
fixed  points  shall  be  equal  to  a  given  line. 

The  two  fixed  points  are  the  foci. 

The  curve  may  be  constructed  by  points,  thus :  Let  F  and  F', 
Fig.  36,  be  the  two  foci,  and  YY'  the  given  line,  so  placed  that 
FY  =  FY'. ' 

With  F'  as  a  centre,  and  any  radius  greater  than  F'V,  as  F'O, 
describe  an  arc.  With  F  as  a  centre,  and  a  radius  FM,  equal  to 
F'O  —  YY',  describe  a  second  arc,  intersecting  the  first  in  the 
points  M  and  M'.  These  will  be  points  of  the  required  curv<», 
since 

F'M  -  FM  -  F'O  -  FM  =  YY'  j    also  F'M'  -  FM'  =  VY 
In  the  same  way  any  number  of  points  may  be  determined. 


30  DESCRIPTIVE    GEOMETRY. 

It  is  manifest,  also,  that  if  the  greater  radius  be  used  with  F  as 
a  centre,  another  branch,  NV'N',  exactly  equal  to  MVM',  will  be 
described. 

V  and  V  are  evidently  points  of  the  curve,  since 

F'V  -  FV  =  W  M  FV  -  F'V, 

and  are  ike  vertices  of  the  hyperbola. 

The  point  C,  midway  between  the  foci,  is  the  centre,  and  VV  is 
the  transverse  axis.  A  perpendicular,  DD',  to  the  transverse  axis 
at  the  centre,  is  the  indefinite  conjugate  axis.  It  evidently  does 
not  intersect  the  curve. 


PROJECTION    OF    CURVES. 

60.  If  all  the  points  of  a  curve  be  projected  upon  the  horizontal 
plane,  and  these  projections  be  joined  by  a  line,  this  line  is  the 
horizontal  projection  of  the  curve. 

Likewise,  if  the  vertical  projections  of  all  the  points  of  a  curve 
be  joined  by  a  line,  it  will  be  the  vertical  projection  of  the  curve. 


61.  The  two  projections  of  a  curve  being  given,  the  curve  'will,  in 
general,  be  completely  determined.  For  in  the  same  perpendicular 
to  the  ground  line,  two  points,  one  on  each  projection,  may  be 
assumed,  and  the  corresponding  point  of  the  curve  determined,  as 
in  Art  (8).  Thus  m  and  m',  Fig.  37,  being  assumed  in  a  perpen- 
dicular to  AB,  M  will  be  a  point  of  the  curve,  and  in  the  same  . 
way  every  point  of  tbe  curve  may  in  general  be  determined. 

62.  If  the  plane  of  a  curve  of  single  curvature  is  perpendicular 
to  either  plane  of  projection,  the  projection  of  the  curve  on  that 
plane  will  be  a  right  line,  and  all  of  its  points  will  be  projected 
into  the  trace  of  the  plane  on  this  plane  of  projection. 


DESCRIPTIVE   GEOMETRY.  37 

If  the  plane  of  the  curve  be  perpendicular  to  the  ground  line, 
both  projections  will  be  right  lines,  perpendicular  to  the  ground 
line,  and  the  curve  will  be  undetermined,  as  in  Art.  (15). 

If  the  plane  of  the  curve  be  parallel  to  either  plane  of  projec- 
tion, its  projection  on  that  plane  will  be  equal  to  itself,  since  each 
element  of  the  curve  will  be  projected  into  an  equal  element, 
Art.  (14).  Its  projection  on  the  other  plane  will  be  a  right  line, 
parallel  to  the  ground  line. 

The  projection  of  a  curve  of  double  curvature  can  in  no  case  be 
a  right  line. 

63.  The  points  in  which  a  curve  pierces  either  plane  of  projec- 
tion can  be  found  by  the  same  rule  as  in  Art.  (2*7).  Thus  o,  Fig. 
37,  is  the  point  in  which  the  curve  MN  pierces  H,  and  p'  the 
point  in  which  it  pierces  V. 


TANGENTS    AND    NORMALS    TO    LINES. 

64.  If  a  fight  line  be  drawn  through  any  point  of  a  curve,  as 
M,  Fig.  38,  intersecting  it  in  another  point,  as  M',  and  then  the 
second  point  be  moved  along  the  curve  towards  M,  until  it  co- 
incides with  it,  the  line,  during  the  motion  containing  both  points, 
will  become  tangent  to  the  curve  at  M,  which  is  the  point  of  contact. 

As  when  the  point  M'  becomes  consecutive  with  M,  the  line 
thus  containing  the  element  of  the  Curve  at  M,  Art.  (57),  may,  for 
all  practical  purposes,  be  regarded  as  the  tangent,  we  say  that  a 
right  line  is  tangent  to  another  line,  when  it  contains  two  consecutive 
points  of  that  line. 

If  a  right  line  continually  approaches  a  curve,  and  becomes 
tangent  to  it,  at  an  infinite  distance,  it  is  called  an  asymptote  of 
the  curve. 

Two  curves  are  tangent  to  each  other,  when  they  contain  two 
consecutive  points,  or  have,  at  a  common  point,  a  common  tangent.  \ 

If  a  right  line  is  tangent  to  a  curve  of  tingle  curvature,  it  will 


38  DESCRIPTIVE   GEOMETRY. 

be  contained  in  the  plane  of  the  curve,  for  it  passes  through  two 
points  in  that  plane,  viz.,  the  two  consecutive  points  of  the  curve. 

Also,  if  a  right  line  is  tangent  to  another  right  line,  it  will  coin- 
cide with  it,  as  the  two  lines  have  two  points  in  common. 
.'      The  expression,  " a  tangent  to  a  curve,"  or  "a  tangent,"  will 
hereafter  be   understood  to  mean   a   rectilineal  tangent,   unless 
otherwise  mentioned. 


65.  If  two  lines  are  tangent  in  space,  their  projections  on  the 
same  plane  will  be  tangent  to  each  other.  For  the  projections  of 
the  two  consecutive  points,  common  to  the  two  lines,  will  also  be 
consecutive  points,  common  to  the  projections  of  both  lines,  Art. 
(60). 

The  converse  of  this  is  not  necessarily  true.  But  if  both  the 
horizontal  and  vertical  projections  are  tangent  at  points,  which  are 
the  projections  of  a  common  point  of  the  two  lines,  Art.  (23),  the 
lines  will  be  tangent  in  space  ;  for  the  projecting  perpendiculars, 
at  the  common  consecutive  points,  will  intersect  in  two  consecu- 
tive points  common  to  the  two  lines. 


66.  If  a  right  line  be  drawn  perpendicular  to  a  tangent  at  its  • 
point  of  contact,  as  MO,  Fig.  38,  it  is  a  normal  to  the  curve.     As 
an  infinite  number  of  perpendiculars  can  be  thus  drawn,  all  in  a 
plane  perpendicular  to  MT  at  M,  there  will  be  an  infinite  number 
of  normals  at  the  same  point. 

If  the  curve  be  a  plane  curve,  that  is,  a  curve  of  single  curva- 
ture, the  term  "  normal"  will  be  understood  to  mean  that  normal 
which  is  in  the  plane  of  the  curve,  unless  otherwise  mentioned. 


o7.  If  we  conceive  a  curve  to  be  rolled  on  its  tangent  at  any 
point,  until  each  of  its  elements  in  succession  comes  into  this  tan- 
gent, the  curve  is  said  to  be  rectified;  that  is  a  right  line,  equal  to 
it  in  length,  has  been  found. 


DESCRIPTIVE   GEOMETRY. 


Since  the  tangent  to  a  curve  at  a  point  contains  the  element  of 
the  curve,  the  angle  which  the  curve,  at  this  point,  makes  with 
any  line  or  plane  will  be  the  same  as  that  made  by  the  tangent. 


THE    HELIX. 

68.  If  a  point  be  moved  uniformly  around  a  right  line,  remain- 
ing always  at  the  same  distance  from  it,  and  having  at  the  same 
time  a  uniform  motion  in  the  direction  of  the  line,  it  will  generate 
a  curve  of  double  curvature,  called  a  helix* 

The  right  line  is  the  axis  of  the  curve. 

Since  all  the  points  of  the  curve  are  equally  distant  from  the 
axis,  the  projection  of  the  curve  on  a  plane  perpendicular  to  this 
axis  will  be  the  circumference  of  a  circle. 

Thus  let  ra,  Fig.  39,  be  the  horizontal,  and  m!n'  the  vertical 
projection  of  the  axis,  and  P  the  generating  point,  and  suppose 
that  while  the  point  moves  once  around  the  axis,  it  moves 
through  the  vertical  distance  mV;  prqs  will  be  the  horizontal 
projection  of  the  curve. 

To  determine  the  vertical  projection,  divide  prqs  into  any  num- 
ber of  equal  parts,  as  16,  and  also  the  line  m'n'  into  the  same 
number,  as  in  the  figure.  Through  these  points  of  division  draw 
lines  parallel  to  AB.  Since  the  motion  of  the  point  is  uniform, 
while  it  moves  one-eighth  of  the  way  round  the  axis  it  will  as- 
cend one-eighth  of  the  distance  m'n\  and  be  horizontally  pro- 
jected at  a?,  and  vertically  at  x'.  When  the  point  is  horizontally 
projected  at  r,  it  will  be  vertically  projected  at  r' ;  and  in  the  same 
way  the  points  y \  q',  &c,  may  be  determined,  and  p'r'q's'  will  be 
the  required  vertical  projection. 


69.  It  is  evident  from  the  nature  of  the  motion  of  the  genera- 
ting point,  that  in  generating  any  two  equal  portions  of  the  curve, 
it  ascends  the  same  vertical  distance  ;  that  is>  any  two  elementary 


40  DESCRIPTIVE   GEOMETRY. 

arcs  of  the  curve  will  make  equal  angles  with  the  horizontal  plane* 
Thus,  if  CD  (a),  Fig.  39,  be  any  element  of  the  curve,  the  angle 
which  it  makes  with  the  horizontal  plane  will  be  DO,  cr  the 
angle  at  the  base  of  a  right-angled  triangle  of  which  Ce  ==  cd,  Fig. 
39,  is  the  base  and  De  the  altitude.  But  from  the  nature  of  the 
motion,  Ce  is  to  De  as  any  arc  px  is  to  the  corresponding  ascent 
x"x'.  Hence,  if  we  rectify  the  arc  xp,  Art.  (67),  and  with  this  as 
a  base  construct  a  right-angled  triangle,  having  x'x"  for  its  alti- 
tude, the  angle  at  the  base  will  be  the  angle  which  the  are,  or  its 
tangent  at  any  point,  makes  with  the  horizontal  plane.  There- 
fore, to  draw  a  tangent  at  any  point  as  X,  we  draw  xz  tangent  to 
the  circle  pxr  at  x  j  it  will  be  the  horizontal  projection  of  the  re- 
quired tangent.  On  this,  from  x,  lay  off  the  rectified  arc  xp  to  z  ; 
v  will  be  the  point  where  the  tangent  pierces  H,  and  z'x'  will  be 
its  vertical  projection. 

Since  the  angle  which  a  tangent  to  the  helix  makes  with  the 
horizontal  plane  is  constant,  and  since  each  element  of  the  curve 
is  equal  to  the  hypothenuse  of  a  right-angled  triangle  of  which 
the  base  is  its  horizontal  projection,  the  angle  at  the  base,  the  con- 
stant angle,  and  the  altitude,  the  ascent  of  the  point  while  genera- 
ting the  element ;  it  follows,  that  when  the  helix  is  rolled  out  on 
its  tangent,  the  sum  of  the  elements,  or  length  of  any  portion  of 
the  curve,  will  be  equal  to  the  hypothenuse  of  a  right-angled  tri- 
angle, of  which  the  b&se  is  its  .horizontal  projection  rectified,  and 
altitude,  the  ascent  of  the  generating  point  while  generating  the 
portion  considered.  Thus  the  length  of  the  arc  pX  is  equal  to 
the  length  of  the  portion  of  the  tangent  ZX. 


GENERATION    AND1    CLASSIFICATION    OF    SURFACES. 

70.  A  surface  may  be  generated  by  the  continued  motion  of  a 
line.  The  moving  line  is  the  generatrix  of  the  surface ;  and  the 
different  positions  of  the  generatrix  are  the  elements. 

If  the  generatrix  be  taken  in  any  position,  and  then  be  moved 


DESCRIPTIVE   GEOMETRY.  41 

to  its  next  position  on  the  surface,  these  two  positioi  s  are  consecu* 
tive  positions  of  the  generatrix,  or  consecutive  elements  of  the  sur- 
face, and  may  practically  be  regarded  as  one  element 


<T.  The  form  of  the  generatrix,  and  the  law  which  directs  its 
motion,  determine  the  nature  and  class  of  the  surface. 

Surfaces  may  be  divided  into  two  general  classes. 

First.  Those  which  can  be  generated  by  right  lines ;  or  whkh 
have  rectilinear  elements. 

Second-  Those  which  can  only  be  generated  by  curves,  and  which 
can  have  no  rectilinear  elements.  These  are  Double  curved  sur- 
faces. 

Those  which  can  be  generated  by  right  lines  are : 

First.  Planes,  which  may  be  generated  by  a  right  line  moving 
so  as  to  touch  another  right  line,  having  all  its  positions  parallel 
to  its  first  position. 

Second.  Single  curved  surfaces,  which  may  be  generated 
by  a  right  line,  moving  so  that  any  two  of  its  consecutive  positioiut 
shall  be  in  t/ie  same  plane. 

Third*  Warped  surfaces,  which  may  be  generated  by  a  right 
line  moving  so  that  no  two  of  its  consecutive  positions  shall  be  in, 
the  same  plane. 


12.  Single  curved  surfaces  are  of  three  kinds. 

I.  Those  in  which  all  tfie  positions  of  the  rectilinear  generatrix 
are  parallel. 

II.  Those  in  which  all  the  positions  of  the  rectilinear  generatrix 
intersect  in  a  common  point 

IIL  Those  in  which  the  consecutive  positions  of  the  rectilinear 
generatrix  intersect  two  and  two,  no  three  positions  intersecting  in 
a  common  point 


42  DESCRIPTIVE    GEOMETRY. 


CYLINDRICAL    SURFACES,    OR    CYLINDERS. 

73.  Single  curved  surfaces  of  the  first  kind  are  Cylindrical  sur- 
faces,  or  Cylinders.  Every  cylinder  may  be  generated  by  moving 
a  right  line  so  as  to  touch  a  curve,  and  have  all  its  positions 
parallel. 

The  moving  line  is  the  rectilinear  generatrix.  The  curve  is  the 
directrix.  The  different  positions  of  the  generatrix  are  the  recti- 
linear elements  of  the  surface. 

Thus,  Fig.  40,  if  the  right  line  MN  be  moved  along  the  curve 
mlo,  having  all  its  positions  parallel  to  its  first  position,  it  will 
generate  a  cylinder. 

It*  the  cylinder  be  intersected  by  any  plane  not  parallel  to  the 
rectilinear  elements,  the  curve  of  intersection  may  be  taken  as  a 
directrix,  and  any  rectilinear  element  as  the  generatrix,  and  the 
surface  be  re-generated.  This  curve  of  intersection  may  also  be 
the  base  of  the  cylinder. 

The  intersection  of  the  cylinder  by  the  horizontal  plane  is 
usually  taken  as  the  base.  If  this  base  have  a  centre,  the  right 
line  through  it,  parallel  to  the  rectilinear  elements,  is  the  axis  of 
the  cylinder. 

A  definite  portion  of  the  surface  included  by  two  parallel  planes 
is  sometimes  considered ;  in  which  case  the  lower  curve  of  inter- 
section is  the  lower  base,  and  the  other  the  upper  base. 

Cylinders  are  distinguished  by  the  name  of  their  bases ;  as  a 
cylinder  with  a  circular  base  ;  a  cylinder  with  an  elliptical  base. 

If  the  rectilinear  elements  are  perpendicular  to  the  plane  of  the 
base,  the  cylinder  is  a  right  cylinder,  and  the  base  a  right  section. 

A  cylinder  may  also  be  generated  by  moving  the  curvilinear 
directrix,  as  a  generatrix,  along  any  one  of  the  rectilinear  ele 
,  ments,  as  a  directrix,  the  curve  remaining  always  parallel  to  it 
first  position. 


DESCRIPTIVE   GEOMETRY.  43 

If  the  curvilinear  directrix  be  changed  to  a  right  line,  the  cyl- 
inder becomes  a  plane. 

It  is  manifest  that  if  a  plane  parallel  to  the  rectilinear  elements 
intersects  the  cylinder,  the  lines  of  intersection  will  be  rectilinear 
elements,  which  will  intersect  the  base. 


74.  It  will  be  seen  that  the  projecting  lines  of  the  different 
points  of  a  curve,  Art.  (60),  form  a  right  cylinder,  the  base  of 
which,  in  the  plane  of  projection,  is  the  projection  of  the  curve. 

These  cylinders  are  respectively  the  horizontal  and  vertical  pro- 
jecting cylinders  of  the  curve,  and  by  their  intersection  determine 
the  curve. 


75.  A  cylinder  is  represented  by  projecting  one  or  more  of  the 
curves  of  its  surface,  and  its  principal  rectilinear  elements. 

When  these  elements  are  not  parallel  to  the  horizontal  plane,  it 
is  usually  represented  thus:  Draw  the  base,  as  mlo,  Fig.  40,  in 
the  horizontal  plane.  Tangent  to  this,  draw  right  lines  Ix  and  kr, 
parallel  to  the  horizontal  projection  of  the  generatrix ;  these  will 
be  the  horizontal  projections  of  the  extreme  rectilinear  elements, 
as  seen  from  the  point  of  sight,  thus  forming  the  horizontal  pro- 
jection of  the  cylinder.  Draw  tangents  to  the  base,  perpendicular 
to  the  ground  line,  as  mm',  oo' ;  through  the  points  m'  and  o', 
draw  lines  m'nf  and  oV,  parallel  to  the  vertical  projection  of  the 
generatrix,  thus  forming  the  vertical  projection  of  the  cylinder; 
m'o'  being  the  vertical  projection  of  the  base. 


76.  To  assume  a  point  of  the  surface,  we  first  assume  one  of  its 
projections,  as  the  horizontal.  Thrpugh  this  point,  erect  a  per 
pcndicular  to  the  horizontal  plane.  It  will  pierce  the  surface  in 
the  only  points  which  can  be  horizontally  projected  at  the  point 
taken.     Through  this  perpendicular,  pass  a  plane  parallel  to  the 


44  DESCRIPTIVE    GEOMETRY. 

rectilinear  elements;  it  will  intersect  the  cylinder  in  elements, 
Art.  (IS),  which  will  be  intersected  by  the  perpendicular  in  the 
required  points. 

Construction.  Let  p,  Fig.  40,  be  the  horizontal  projection  as- 
sumed. Through  p,  draw  pq  parallel  to  Ix ;  it  will  be  the  hori- 
zontal trace  of  the  auxiliary  plane.  This  plane  intersects  the 
cylinder  in  two  elements ;  one  of  which  pierces  H  at  q,  and  the 
other  at  u  ;  and  q'y',  and  u'z' ',  will  be  the  vertical  projections 
of  these  elements,  v'p'  the  vertical  projection  of  the  perpendicu- 
lar, and  p'  and  p"  the  vertical  projections  of  the  two  points  of 
the  surface,  horizontally  projected  at  p. 

To  assume  a  rectilinear  element,  we  have  simply  to  draw  a  line 
parallel  to  the  rectilinear  generatrix,  through  any  point  of  the 
base,  or  of  the  surface. 


CONICAL    SURFACES,    OR    CONES. 

77.  Single  curved  surfaces  of  the  second  kind,  are  Conical  sur- 
faces, or  Cones. 

Every  cone  may  be  generated  by  moving  a  right  line  so  as  con- 
tinually to  touch  a  given  curve,  and  pass  through  a  given  point 
not  in  the  plane  of  the  curve. 

The  moving  line  is  the  rectilinear  generatrix ;  the  curve,  the 
directrix;  the  given  point,  the  vertex  of  the  cone;  and  the  differ- 
ent positions  of  the  generatrix,  the  rectilinear  elements. 

The  generatrix  being  indefinite  in  length,  will  generate  two 
parts  of  the  surface,  on  different  sides  of  the  vertex,  which  are 
called  nappes  ;   one,  the  upper,  the  other,  the  lower  nappe. 

Thus,  if  the  right  line  MS,  Fig.  41,  move  along  the  curve  mlo 
and  continually  pa<s  through  S,  it  will  generate  a  cone. 

If  the  cone  be  intersected  J>y  any  plane  not  passing  through 
the  vertex,  the  curve  of  intersection  may  be  taken  as  a  directrix, 
and  any  rectilinear  element  as  a,  generatrix,  and  the  cone  be  re- 
generated.    This  curve  of  intersection  may  also  be  the  base  of  tho 


DESCRIPTIVE    GKOMETRY.  45 

cone.  The  intersection  of  the  cone  by  the  horizontal  plane,  is 
usually  taken  as  the  base. 

If  a  definite  portion  of  the  cone  included  by  two  parallel  plane3 
is  considered,  it  is  called  a  frustum  of  a  cone  ;  one  of  the  limit- 
ing curves  being  the  lower,  and  the  other  the  upper  base  of  the 
fi  ustum. 

Cones  are  distinguished  by  the  names  of  their  bases ;  as  a  cone 
with  a  circular  base ;  a  cone  with  a  parabolic  base,  &c. 

If  the  rectilinear  elements  all  make  the  same  angle  with  a  right 
line  passing  through  the  vertex,  the  cone  is  a  right  cone,  the  right 
line  being  its  axis. 

A  cone  may  also  be  generated  by  moving  a  curve  so  as  con- 
tinually to  touch  a  right  line,  and  change  its  size  according  to  a 
proper  law. 

If  the  curvilinear  directrix  of  a  cone  be  changed  to  a  right 
line,  or  if  the  vertex  be  taken  in  the  plane  of  the  curve,  the  cone 
will  become  a  plane. 

If  the  verttx  be  removed  to  an  infinite  distance,  the  cone  will 
evidently  become  a  cylinder. 

If  a  cone  be  intersected  by  a  plane  through  the  vertex,  the 
lines  of  intersection  will  be  rectilinear  elements,  intersecting  the 
base. 

*78.  A  cone  is  represented  by  projecting  the  vertex,  one  of  the 
curves  on  its  surface,  and  its  principal  rectilinear  elements.  Thus, 
let  S,  Fig.  41,  be  the  vertex.  Draw  the  base,  mlo,  in  the  horizon- 
tal plane,  and  tangents  to  this  base  through  s,  as  si  and  sk ;  thus 
forming  the  horizontal  projection  of  the  cone.  Draw  tangents  to 
the  base,  perpendicular  to  the  ground  line,  as  mm\  oo' ;  and 
through  m'  and  o',  draw  the  right  lines  m'sf  and  o's',  thus  foim- 
ing  the  vertical  projection  of  the  cone. 


*79.    To  assume  a  point  of  the  surface,  we  first  assume  one  o 
its  projections,  as  the  horizontal.     Through  this  erect  a  perpen 


4:6  DESCRIPTIVE   GEOMETRY. 

dicular  to  the  horizontal  plane ;  it  will  pierce  the  surface  in  the 
only  points  which  can  be  horizontally  projected  at  the  point 
taken.  Through  this  perpendicular  and  the  vertex  pass  a  plane. 
It  will  intersect  the  cone  in  elements  which  will  be  intersected  by 
he  perpendicular  in  the  required  points. 

Construction.  Let  p  be  the  horizontal  projection.  Draw  ps. 
It  will  be  the  horizontal  trace  of  the  auxiliary  plane,  which  inter- 
sects the  cone  in  two  elements ;  one  of  which  pierces  H  at  q,  and 
the  other  at  r,  and  q's'  and  r's'  are  the  vertical  projections  of 
these  elements,  and  p'  and  p"  are  the  vertical  projections  of  the 
two  points  of  the  surface. 

To  assume  a  rectilinear  element,  we  have  simply  to  draw 
through  any  point  of  the  base,  or  of  the  surface,  a  right  line  to 
the  vertex. 


80.  Single  curved  surfaces  of  the  third  kind,  may  be  generated 
by  drawing  a  system  of  tangents  to  any  curve  of  double  curva- 
ture. These  tangents  will  evidently  be  rectilinear  elements  of  a 
single  curved  surface.  For  if  we  conceive  a  series  of  consecutive 
points  of  a  curve  of  double  curvature,  as  a,  b,  c,  d,  &c,  the  tan- 
gent which  contains  a  and  6,  Art.  (64),  is  intersected  by  the  one 
which  contains  b  and  c,  at  b ;  that  which  contains  b  and  c,  by  the 
one  which  contains  c  and  d,  at  c ;  and  so  on,  each  tangent  inter- 
secting the  preceding  consecutive  one,  but  not  the  others,  since  no 
two  elements  of  the  curve,  not  consecutive,  are,  in  general,  in  the 
same  plane,  Art.  (58). 


81.  If  the  curve  to  which  the  tangents  are  drawn,  is  a  helix, 
Art.  (68),  the  surface  may  be  represented  thus :  Let  pxy,  Fig. 
42,  be  the  horizontal,  and  p'x'y'  the  vertical  projection  of  the 
helical  directrix.  Since  the  rectilinear  elements  are  all  tangent  to 
this  directrix,  any  one  may  be  assumed,  as  in  Art.  (69) ;  hence 
XZ,  YU,  &c,  are  elements  of  the  surface. 


DESCRIPTIVE    GEOMETRY.  47 

A  point  of  the  surface  may  be  assumed  b\  taking  a  point  on 
any  assumed  element. 

These  elements  pierce  H  in  the  points  z,  u,  v,  &c,  and  zuvw  is 
the  curve  in  which  the  surface  is  intersected  by  the  horizontal 
plane,  and  may  be  regarded  as  the  base  of  the  surface;  and  it  is 
evident  that  if  the  surface  be  intersected  by  any  plane  parallel  to 
this  base,  the  curve  of  intersection  will  be  equal  to  the  base. 


WAKPED    SURFACES    WITH    A    PLANE    DIRECTER. 

82.  There  is  a  great  variety  of  warped  surfaces,  differing  from 
each  other  in  their  mode  of  generation  and  properties. 

The  most  simple  are  those  which  may  be  generated  by  a  right 
line  generatrix,  moving  so  as  to  touch  two  other  lines  as  direc- 
trices, and  parallel  to  a  given  plane,  called  a  plane  directer. 

Such  surfaces  are  warped  surfaces,  with  two  linear  directrices 
and  a  plane  directer. 

They,  as  all  other  warped  surfaces,  may  be  represented  by  pro- 
jecting one  or  more  curves  of  the  surface,  and  the  principal  rec- 
tilinear elements. 


83.  The  directrices  and  plane  directer  being  given,  a  recti- 
linear element,  passing  through  any  point  of  either  directrix,  may 
be  determined,  by  passing  a  plane  through  this  point,  parallel  to 
the  plane  directer,  and  finding  the  point  in  which  this  plane  cuts 
the  other  directrix,  and  joining  this  with  the  given  point. 

Construction.  Let  BIN  and  PQ,  Fig.  43,  be  any  two  linear 
directrices,  tTt'  the  plane  directer,  and  O  any  point  of  the  first 
directrix. 

Assume  any  line  of  the  plane  directer,  as  CD,  Art  (30),  and 
through  the  different  points  of  this  line,  draw  right  lines  SE,  SF, 
&c,  to  any  point,  as  s  of  *T.  Through  0  draw  a  system  of  lines 
OR,  OY,  &c,  parallel   respectively  to  SE,  SF,  <fcc.     These  will 


48  DESCRIPTIVE    GEOMETRY. 

form  a  plane  through  0,  parallel  to  tTt',  and  pierce  the  horizontal 
projecting  cylinder  of  PQ,  in  the  points  R,  Y,  W,  &c.  These 
points  being  joined,  will  form  the  curve  RW,  which  intersects 
PQ  in  X,  and  this  is  the  point  in  which  the  auxiliary  plane  cuts 
the  directrix  PQ.  OX  will  then  be  the  required  element.  If  the 
curve  r'w'  should  intersect  p'q'  in  more  than  one  point,  two  or 
more  elements  passing  through  O  would  thus  be  determined. 


84.  If  an  element  be  required  parallel  to  a  given  right  line,  this 
line  being  in  the  plane  directer,  or  parallel  to  it,  we  draw  through 
the  different  points  of  either  directrix,  lines  parallel  to  the  given 
line.  These  form  the  rectilinear  elements  of  a  cylinder,  parallel 
to  the  given  line.  If  the  points  in  which  the  second  directrix 
pierces  this  cylinder  be  found,  and  lines  be  drawn  through  them 
parallel  to  the  given  line,  each  will  touch  both  directrices,  and  be 
an  element  required. 

Construction.  Let  the  surface  be  given  as  in  the  preceding 
Article,  and  let  FS,  Fig.  44,  be  the  given  line.  Through  the 
points  O,  K,  L,  &c,  draw  OX,  KY,  LZ,  &c,  parallel  to  FS. 
These  lines  pierce  the  horizontal  projecting  cylinder  of  the  direc- 
trix PQ.  in  the  points  X,  Y,  Z,  &c,  which,  being  joined,  form  the 
curve  XY,  intersecting  PQ  in  W.  Through  W  draw  WR,  par- 
allel to  FS  ;  it  is  a  required  element,  and  there  may  be  two  or 
more  as  in  the  preceding  Article. 


85.  A  warped  surface,  with  a  plane  directer,  having  one  direc- 
trix a  right  line  and  the  other  a  curve,  is  called  a  Conoid.  The 
elements  of  the  conoid  pass  through  all  the  points  of  the  recti- 
linear directrix,  instead  of  a  single  point,  as  in  the  cone. 

If  the  rectilinear  directrix  is  perpendicular  to  the  plane  directer, 
it  is  a  right  conoid,  and  this  directrix  is  the  line  of  striction. 


DESCRIPTIVE   GEOMETRY.  49 


THE    HYPERBOLIC    PARABOLOID. 

86.  A  warped  surface  with  a  plane  directer  and  two  rectilinear 
directrices,  is  a  Hyperbolic  Paraboloid,  since  its  intersection  by  a 
plane  may  be  proved  to  be  either  an  hyperbola  or  parabola. 

Its  rectilinear  elements  may  be  constructed  by  the  principles  in 
Arts.  (83  and  84).  In  the  first  case,  two  right  lines  through  the 
given  point  will  determine  the  auxiliary  plane,  and  the  point  in 
which  it  is  pierced  by  the  second  directrix  may  be  determined  at 
once,  as  in  Art.  (41).  In  the  second  case,  the  cylinder  becomes  a 
plane,  and  the  point  in  which  it  is  pierced  by  the  second  directrix 
is  also  determined  as  in  Art.  (41). 


87.  The  rectilinear  elements  of  a  hyperbolic  paraboloid  divide  the 
directrices  proportionally ;  for  these  elements  are  in  a  system  of 
planes  parallel  to  the  plane  directer  and  to  each  other,  and  these 
planes  divide  the  directrices  into  proportional  parts  at  the  points 
where  they  are  intersected  by  the  elements.  (Davies'  Legendre, 
Book  vi.,  Prop,  xv.) 

Conversely.  If  two  right  lines  be  divided  into  any  number  of 
proportional  parts,  the  right  lines  joining  the  corresponding  points 
of  division  will  lie  in  a  system  of  parallel  planes,  and  be  elements 
of  an  hyperbolic  paraboloid,  the  plane  directer  of  which  is  parallel 
to  any  two  of  these  dividing  lines. 

Thus,  let  MN  and  OP,  Fig.  45,  be  any  two  rectilinear  di- 
rectrices. Take  any  distance,  as  ml,  and  lay  it  off  on  mn  any 
number  of  times,  as  ml,  Ik,  kg,  <fec.  Take  also  any  distance,  as 
pr,  and  lay  it  off  on  po  any  number  of  times,  as  pr,  rs,  su,  go,  &c. 
Join  the  corresponding  points  of  division  by  right  lines,  Ir,  ks,  gu, 
&c. ;  these  will  be  the  horizontal  projections  of  rectilinear  ele- 
ments of  the  surface.  Through  m,  I,  k,  &c,  and  pf  r,  6-,  &a,  erect 
perpendiculars  to  AB,  to  m\  V ,  k',  &c,  and  p\  r\  s',  &c,  and  join 
4 


50  DESCRIPTIVE   GEOMETBY. 

he  corresponding  points  by  the  right  lines  Vr\  k's\  g'u\  &c., 
these  will  be  the  vertical  projections  of  the  elements. 


88.  To  assume  any  point  on  this  s:irface,  and  in  general,  on  any 
Warped  surface,  we  first  assume  its  horizontal  projection,  and  at 
this,  erect  a  perpendicular  to  the  horizontal  plane.  Through  this 
perpendicular  pass  a  plane  ;  it  will  intersect  the  rectilinear  el* 
ements  in  points  which,  joined,  will  give  a  line  of  the  surface, 
and  the  points  in  which  this  line  intersects  the  perpendicular  will 
be  the  required  points  on  the  surface. 

Let  the  construction  be  made  upon  either  of  the  figures,  43,  44, 
or  45. 

89.  If  any  two  rectilinear  elements  of  an  hyperbolic  paraboloid  be 
taken  as  directrices,  with  a  plane  directer  parallel  to  the  first  di- 
rectrices, and  a  surface  be  thus  generated,  it  will  be  identical  with 
the  first  surface. 

To  prove  this,  we  have  only  to  prove  that  any  point  of  an  ele* 
ment  of  the  second  generation  is  also  a  point  of  the  first* 

Thus,  let  MN  and  OP,  Fig.  46,  be  the  directrices  of  the  first 
generation,  and  NO  and  MP  any  two  rectilinear  elements. 
Through  M  draw  MW  parallel  to  NO.  The  plane  WMP  will  be 
parallel  to  the  plane  directer  of  the  first  generation,  and  may  be 
taken  for  it. 

Let  NO  and  MP  be  taken  as  the  new  directrices,  and  let  ST  be 
an  element  of  the  second  generation,  the  plane  directer  being 
parallel  to  MN  and  OP»  Through  U,  any  point  of  ST,  pass  a 
plane  parallel  to  WMP,  cutting  the  directrices  MN  and  OP  in  Q 
and  R.  Join  QR,  it  will  be  an  element  of  the  first  generation, 
Art.  (83).  Draw  N»  and  Qq  parallel  to  ST,  piercing  the  plane 
WMP  in  n  and  q.  Also  draw  Oo  and  Rr  parallel  to  ST,  piercing 
WMP  in  o  and  r;  and  draw  no,  intersecting  MP  in  T,  since  Nm, 
ST,  and  Oo  are  in  the  same  plane.  M,  q,  and  n  will  be  in  the 
same  right  line,  as  also  P,  r,  and  o ;  and  since  MN  and  Nrc,  inter- 


DESCRIPTIVE   GEOMETRY  51 

secting  at  N,  are  parallel  to  the  plane  direfcter  of  the  second  gen- 
eration, their  plane  will  be  parallel  to  it,  as  also  the  plane  of  PO 
and  Oo ;    hence,  these  planes  being  parallel,  their  intersections, 
M»  and  Po,  with  the  plane  WMP,  will  be  parallel.     Draw  qr. 
Since  Qq  and  Nn  are  parallel,  we  have 


Also, 

But  Art.  (87), 

hence, 


MQ  :  QN   :  :  Uq   :  qn. 
PR  :  BO  :  :  Pr  :  ro> 

MQ  :  QN  :  :  PR  :  RO; 

Mq    :   qn   :  :   Tr   :   ro ; 


and  the  right  line,  qr,  must  pass  through  T,  and  the. plane  of  the 
three  parallels,  Qq,  ST,  and  Rr,  contains  the  element  QR,  which 
must  therefore  intersect  ST  at  U.  Hence,  any  point  of  a  rec- 
tilinear element  of  the  second  generation  is  also  a  point  of  an 
element  of  the  first  generation,  and  the  two  surfaces  are  identical. 
It  follows  from  this,  that  through  any  point  of  an  hyperbolic  parabo* 
loid  two  rectilinear  elements  can  always  be  drawn. 


WARPED    SURFACES    WITH   THREE    LINEAR    DIRECTRICES. 

90.  A  second  class  of  warped  surfaces  consists  of  those  which 
may  be  generated  by  moving  a  right  line  so  as  to  touch  three 
lines  as  directrices ;  or  warped  surfaces  with  three  linear  directrices. 

A  rectilinear  element  of  this  class  of  surfaces  passing  through  a 
given  point-  on  one  of  the  directrices,  may  be  found  by  drawing 
through  this  point  a  system  of  right  lines  intersecting  rather  of 
the  other  directrices ;  these  form  the  surface  of  a  cone  which  wil) 


52  DESCRIPTIVE   GEOMETRY. 

be  pierced  by  the  third  directrix  in  one  or  more  points,  through 
which  and  the  given  point  right  lines  being  drawn  will  touch  the 
three  directrices,  and  be  required  elements. 

Construction.  Let  MN,  OP,  and  QR,  Fig.  47,  be  any  three 
linear  directrices,  and  M  a  gives  point  on  the  first.  Through  M 
draw  the  lines  MO,  MS,  MP,  &c,  intersecting  OP  in  O,  S,  P,  &c. 
They  pierce  the  horizontal  projecting  cylinder  of  QR  in  X,  Y,  Z, 
tfcc,  forming  a  curve,  XYZ,  which  intersects  QR  in  U,  the  point 
in  which  QR  pierces  the  surface  of  the  auxiliary  cone.  MU  is 
then  a  required  element,  two  or  more  of  which  would  be  deter- 
mined if  XZ  should  intersect  QR  in  more  than  one  point. 


91.  The  simplest  of  the  above  class  of  surfaces  is  a  surface 
which  may  be  generated  by  moving  a  right  line  so  as  to  touch  three 
rectilinear  directrices.  It  is  an  Hyperboloid  of  one  nappe,  as  many 
of  its  intersections  by  planes  are  hyperbolas,  while  the  surface 
itself  is  unbroken. 

The  construction  of  the  preceding  Article  is  much  simplified 
for  this  surface,  as  the  auxiliary  cone  becomes  a  plane;  and  the 
point  in  which  this  plane  is  pierced  by  the  third  directrix  is 
found  as  in  Art.  (40),  or  (41). 


THE    HELICOID. 

92.  If  a  right  line  be  moved  uniformly  along  another  right  line, 
as  a  directrix,  always  making  the  same  angle  with  it,  and  at  the 
same  time  having  a  uniform  angular  motion  around  it,  a  warped 
surface  will  be  generated,  eailed  a  Hdicoid. 

The  rectilinear  directrix  is  the  axis  of  the  surface.     Thus,  Fig 
48,  let  the  horizontal  plane  be  taken  perpendicular  to  the  axis, 
being  its  horizontal,  and  o'n'  its  vertical  projection,  and  let  OP  be 
the  generatrix,  parallel  to  the  vertical  plane. 


DESCRIPTIVE    GEOMETRY.  53 

It  is  evident  from  the  nature  of  the  motion  of  the  generatrix, 
that  each  of  its  points  will  generate  a  helix,  Art.  (68).  That  gen- 
erated by  the  point  P,  constructed  as  in  Art.  (68),  will  be  hori- 
zontally projected  in  prq,  and  vertically  in  p'r'q'. 

To  assume  a  rectilinear  element  of  the  surface,  we  first  assume 
its  horizontal  projection,  as  ox.  Through  x  erect  the  perpendicu- 
lar xx',  and  from  o'  lay  off  the  distance  o'o",  equal  to  x'x"  \  o"x' 
will  be  the  required  vertical  projection.  In  the  same  way,  the 
element  (oy,  o'"y')  may  be  assumed. 

To  assume  any  'point  on  the  surface,  we  first  assume  a  rectilinear 
element,  and  then  take  any  point  in  this  element,  as  M. 

The  elements  pierce  the  horizontal  plane  in  the  points  p,  u,  w, 
&c,  and  the  curve  puw  is  its  intersection  by  the  surface. 

The  surface  is  manifestly  a  warped  surface,  since,  from  the  na- 
ture of  its  generation,  no  two  consecutive  positions  of  the  genera- 
trix can  be  parallel  or  intersect. 

This  surface  is  an  important  one,  as  it  forms  the  curved  surface 
of  the  thread  of  the  ordinary  screw. 

If  the  generatrix  is  perpendicular  to  the  axis,  the  helicoid  be- 
comes a  particular  case  of  the  right  conoid,  Art.  (85),  the  hori- 
zontal plane  being  the  plane  directer,  and  any  helix  the  curvi- 
linear directrix. 


93.  Other  varieties  of  Avarped  surfaces  may  be  generated  by 
moving  a  right  line  so  as  to  touch  two  lines,  having  its  different' 
positions,  in  succession,  parallel  to  the  different  rectilinear  el- 
ements of  a  cone:  By  moving  it  parallel  to  a  given  plane,  so  as 
to  touch,  two  surfaces,  or  one  surface  and  a  line ;  or  so  as  to 
touch  three  surfaces — two  surfaces  and  a  line — one  surface  and 
two  lines ;  and  in  general,  so  as  to  fulfil  any  three  reasonable 
conditions. 

It  should  be  remarked,  that  in  all  these  cases  the  directrices 
should  be  so  chosen  that  the  surface  generated  will  be  neither  a, 
plane  nor  single  curved  surface.     Also,  that  these  surfaces,  as  all 


54  DESCRIPTIVE   GEOMETRY. 

others,  may  be  generated  by  curves  moved  in  accordance  with  a 
law  peculiar  to  each  variety. 


94.  If  a  curve  be  moved  in  any  way  so  as  not  to  generate  a 
surface  of  either  of  the  above  classes,  it  will  generate  a  double 
curved  surface,  the  simplest  variety  of  which  is  the  surface  of  a 
sphere. 


95.  An  examination  of  the  modes  of  generating  surfaces,  above 
described,  will  indicate  the  following  test  for  ascertaining  the 
general  class  to  which  any  given  surface  belongs. 

1.  If  a  straight  ruler  can  be  made  to  touch  the  surface  in  every 
direction,  it  must  be  a  plane. 

2.  If  the  ruler  can  be  made  to  touch  a  curved  surface  in  any 
one  direction,  and  then  be  moved  so  as  to  touch  in  a  position 
very  near  to  the  first,  and  the  two  right  lines  thus  indicated  are 
parallel  or  intersect,  the  surface  must  be  single  curved. 

3.  If  the  lines  thus  indicated,  neither  intersect  nor  are  parallel, 
the  surface  must  be  warped. 

4.  If  the  ruler  cannot  be  made  to  touch  the  surface  in  any  di- 
rection, the  surface  must  be  double  curved. 


SURFACES    OF   REVOLUTION. 

96.  A  surface  of  revolution,  is  a  surface  which  may  be  generated 
by  revolving  a  line  about  a  right  line  as  an  axis,  Art.  (IV). 

From  the  nature  of  this  generation,  it  is  evident  that  any  inter- 
section of  such  a  surface  by  a  plane  perpendicular  to  the  axis,  is  • 
the  circumference  of  a  circle.  Hence,  the  surface  may  also  be 
generated  by  moving  the  circumference  of  a  circle  with  its  plane 
perpendicular  to  the  axis,  and  centre  in  the  axis,  and  whose 
radius  changes  in  accordance  with  a  prescribed  law. 


DESCRIPTIVE   GEOMETRY.  55 

If  the  surface  be  intersected  by  a  plane  passing  through  the 
axis,  the  line,  of  intersection  is  a  meridian  line,  and  the  plane  a 
meridian  plane ;  and  it  is  also  evident  that  all  meridian  lines  of 
the  same  surface  are  equal,  and  that  the  surface  may  be  generated 
by  revolving  anv  one  of  these  meridian  lines  about  the  axis. 


97.  If  two  surfaces  of  revolution,  having  a  common  axis,  inter- 
sect, the  line  of  intersection  must  be  the  circumference  of  a  circle, 
whose  plane  is  perpendicular  to  the  axis,. and  centre  in  the  axis. 
For  if  a  plane  be  passed  through  any  point  of  the  intersection  and 
the  common  axis,  it  will  cut  from  each  surface  a  meridian  line, 
Art.  (96),  and  these  meridian  lines  will  have  the  point  in  com- 
mon. If  these  lines  be  revolved  about  the  common  axis,  each 
will  generate  the  surface  to  which  it  belongs,  while  the  common 
point  will  generate  the  circumference  of  a  circle  common  to  the 
two  surfaces,  and  therefore  their  intersection.  Should  the  me- 
ridian lines  intersect  in  more  than  one  point,  the  surfaces  will 
intersect  in  two  or  more  circumferences. 


98.  The  simplest  curved  surface  of  revolution  is  that  which 
may  be  generated  by  a  right  line  revolving  about  another  right 
line  to  which  it  is  parallel.  This  is  evidently  a  cylindrical  sur- 
face, Art.  (73),  and  if  the  plane  of  the  base  be  perpendicular  to 
the  axis,  it  is  a.  right  cylinder  with  a  circular  base. 

If  a  right  line  be  revolved  about  another  right  line  which  it 
intersects,  it  will  generate  a  conical  surface,  Art.  (77),  which  is 
evidently  a  right  cone,  the  axis  being  the  line  with  which  the 
rectilinear  elements  make  equal  angles. 

These  are  the  only  two  single  curved  surfaces  of  revolution. 


56  DESCRIPTIVE   GEOMETRY. 


THE    HYPERBOLOID    OF   REVOLUTION    OF   OKE   HAPPE. 

99.  If  a  right  line  be  revolved  about  another  right  line,  not  in 
the  same  plane  with  it,  it  will  generate  a  warped  surface,  which 
is  an  Hyperboloid  of  revolution  of  one  nappe,  Art.  (91). 

To  prove  this,  let  us  take  the  horizontal  plane  perpendicular  to 
the  axis,  and  the  vertical  plane  parallel  to  the  generatrix  in  its 
first  position,  and  let  c,  Fig.  49,  be  the  horizontal,  and  c'm'  the 
vertical  projection  of  the  axis,  and  MP  the  generatrix :  cm  will 
be  the  horizontal,  and  m'  the  vertical  projection  of  the  shortest 
distance  between  these  two  lines,  Art.  (55). 

As  MP  revolves  about  the  axis,  CM  will  remain  perpendicular 
to  it,  and  M  will  describe  a  circumference  which  is  horizontally 
projected  in  mxy1  and  vertically  in  y'x' ;  and  as  CM  is  horizontal, 
its  horizontal  projection  will  be  perpendicular  to  the  horizontal 
projection  of  MP,  in  all  of  its  positions,  Art.  (36),  and  remain  of 
the  same  length  ;  hence  the  horizontal  projection  of  MP,  in  any 
position,  will  be  tangent  to  the  circle  mxy.  No  two  consecutive 
positions  can  therefore  be  parallel.  Neither  can  they  intersect; 
for  from  the  nature  of  the  motion,  any  two  must  be  separated  at 
any  point  by  the  elementary  arc  of  the  circle  described  by  that 
point.     The  surface  is  therefore  a  warped  surface. 

The  point  P,  in  which  MP  pierces  H,  generates  the  circle  pwq, 
which  may  be  regarded  as  the  base  of  the  surface. 

The  circle  generated  by  M,  is  the  smallest  circle  of  the  surface, 
and  is  the  circle  of  the  gorge. 


100.  To  assume  a  rectilinear  element,  we  take  any  point  in  the 
base,  pivq,  and  through  it  draw  a  tangent  to  xmy,  as  wz ;  this  will 
be  the  horizontal  projection  of  an  element.  Through  z  erect  the 
perpendicular  zz' ;   z'  will  be  the  vertical  projection  of  the  point 


DESCRIPTIVE    GEOMETRY.  57 

in  which  the  element  crosses  the  circle  of  the  gorge,  and  w'z'  will 
bo  the  vertical  projection  of  the  element. 

To  assume  a  point  of  the  surface,  we  first  assume  a  rectilineal 
element,  as  above,  and  then  take  any  point  of  this  element, 
Art.  (22). 


101.  If  through  the  point  M,  a  second  right  line,  as  MQ,  be 
drawn  parallel  to  the  vertical  plane,  and  making  with  the  hori- 
zontal plane  the  same  angle  as  MP,  and  this  line  be  revolved 
about  the  same  axis,  it  will  generate  the  same  surface.  For  if 
any  plane  be  passed  perpendicular  to  the  axis,  as  the  plane  whose 
vertical  trace  is  e'g',  it  will  cut  MP  and  MQ  in  two  points,  E  and 
G,  equally  distant  from  the  axis,  and  these  points  will,  in  the  rev- 
olution of  M  P  and  MQ,  generate  the  same  circumference ;  hence 
the  two  surfaces  must  be  identical.  The  surface  having  two  gen- 
erations by  different  right  lines,  it  follows  that  through  any  point 
of  the  surface  two  rectilinear  elements  can  be  drawn. 


102.  Since  the  points  E  and  G  generate  the  same  circumfer- 
ence, it  follows  that  as  MP  revolves,  MQ  remaining  fixed,  the 
point  E  will,  at  some  time  of  its  motion,  coincide  with  G,  and  the 
generatrix,  MP,  intersect  MQ  in  G.  In  the  same  way  any  other 
point,  as  F,  will  come  into  the  point  K  of  MQ,  giving  another 
element  of  the  first  generation,  intersecting  MQ  at  K ;  and  so  for 
each  of  the  points  of  MP  in  succession.  In  this  case,  Id  will  be 
the  horizontal  projection  of  the  element  of  the  first  generation. 
Hence,  if  the  generatrix  of  either  generation  remain  fixed,  it  will  in- 
tersect all  the  elements  of  tlie  other  generation. 

I£  then,  any  three  elements  of  either  generation  be  taken  as 
iirectrices,  and  an  element  of  the  other  generation  be  moved  so 
as  to  touch  them,  it  will  generate  the  surface.  It  is  therefore  an 
hj/perboloid  of  one  nappe,  Art.  (91). 

An  hyperboloid  of  revolution  of  one  nappe  may  thus  be  gen- 


58  DESCRIPTIVE   GEOMETRY. 

erated,  by  moving  a  right  line  so  as  to  touch  three  other  right  lines, 
equally  distant  from  a  fourth  (the  axis),  the  distances  being 
measured  in  the  same  plane  perpendicular  to  the  fourth,  and  the 
directrices  making  equal  angles  with  this  plane. 


103.  If  the  vertical  plane  is  not  parallel  to  the  generatrix  in  its 
first  position,  the  circle  of  the  gorge  may  be  constructed  thus : 
Let  WZ  be  the  first  position  of  the  generatrix,  and  through  c 
draw  cz  perpendicular  to  we ;  it  will  be  the  horizontal  projection 
of  the  radius  of  the  required  circle.  The  point  of  which  z  is  the 
horizontal  projection,  is  vertically  projected  at  z',  and  mzx  will  be 
the  horizontal,  and  y'x'  the  vertical  projection  of  the  circle  of  the 
gorge.  With  c  as  a  centre,  and  cw  as  a  radius,  describe  the  base 
pwq,  and  the  surface  will  be  fully  represented. 


104.  To  construct  a  meridian  curve  of  this  surface,  we  pass  a 
plane  through  the  axis,  parallel  to  the  vertical  plane.  It  will  in- 
tersect the  horizontal  circles,  generated  by  the  different  points  of 
the  generatrix,  in  points  of  the  required  curve,  which  will  be  ver- 
tically projected  into  a  curve  equal  to  itself,  Art.  (62). 

Thus  the  horizontal  plane,  whose  vertical  trace  is  e'g',  intersects 
the  generatrix  in  E,  and  eo  is  the  horizontal  projection  of  the 
circle  generated  by  this  point,  and  0  is  the  point  in  which  this 
circle  pierces  the  meridian  plane. 

In  the  same  way,  the  points  whose  vertical  projections  are  n'y 
x',  n",  o",  &c,  are  determined.  The  plane  whose  vertical  trace 
is  e"o",  at  the  same  distance  from  y'x'  as  e'o',  evidently  deter- 
mines a  point  o"  at  the  same  distance  from  y'x',  as  o' ;  hence  the 
chord  o'o"  is  bisected  by  y'x',  and  the  curve  o'x'o"  is  symmetrical 
with  the  line  y'x'. 

The  distance  e'o'  equal  to  ce  —  me,  is  the  difference  between 
the  hypothenuse  ce  and  base  me  of  a  right-angled  triangle,  having 
the  altitude  rue.     As  the  point  o'  is  further  removed  from  x',  tho 


DESCRIPTIVE   GEOMETRY.  5J) 

altitude  of  the  corresponding  triangle  remains  the  same,  while  the 
hypothenuse  and  base  both  increase.  If  we  denote  the  altitude 
by  </,  and  the  base  and  hypothenuse  by  b  and  h  respectively,  we 
have 

W  —  62  =  d\       and       h  —  b  = E ; 

h  -\r  b 

from  which  it  is  evident  that  the  difference,  e'o\  continually  di 
minishes  as  the  point  o'  recedes  from  x' ;  that  is,  the  curve  x'n'o 
continually  approaches  the  lines  p'm'  and  q'm',  and  will  touch 
them  at  an  infinite  distance.  These  lines  are  then  asymptotes  to 
the  curve,  Art.  (64). 

If,  now,  any  element  of  the  first  generation,  as  the  one  passing 
through  I,  be  drawn,  it  will  intersect  the  element  of  the  second 
generation,  MQ,  in  R,  and  the  corresponding  element  on  the  op 
posite  side  of  the  circle  of  the  gorge  in  U. 

Since  this  element  has  but  one  point  in  common  with  the  me- 
ridian curve,  and  no  point  of  it,  or  of  the  surface,  can  be  vertically 
projected  on  the  right  of  this  curve,  the  vertical  projection  u'r' 
must  be  tangent  to  the  curve  at  i'.  But  since  ri  is  equal  to  iu, 
r'i'  must  be  equal  to  i'u' ;  or  the  part  of  the  tangent  included 
between  the  asymptotes  it  bisected  at  the  point  of  contact.  This  is 
a  property  peculiar  to  the  hyperbola  (Analyt.  Geo.,  Art.  164), 
and  the  meridian  curve  is  therefore  an  hyperbola;  YX  being 
its  transverse  axis,  and  the  axis  of  the  surface  its  conjugate,  Art. 
(59).  If  this  hyperbola  be  revolved  about  its  conjugate  axis,  it 
will  generate  the  surface,  Art.  (96),  and  hence  its  name. 

This  is  the  only  warped  surface  of  revolution. 


105.  The  most  simple  double  curved  surfaces  of  revolution  are 

I.  A  spherical  surface  or  sphere,  which  may  be  generated  b} 
revolving  a  circumference  about  its  diameter. 

II.  An  ellipsoid  of  revolution,  or  spheroid,  which  may  be  gen- 
erated  by  revolving  an    ellipse    about  eitb er  axis.      When   the 


6*0  DESCRIPTIVE    GKOMETRY. 

ellipse  is  revolved  about  the  transverse  axis,  the  surface  is  a  pro 
late  spheroid  ;  when  about  the  conjugate  axis,  an  oblate  spheroid. 

III.  A  paraboloid  of  revolution,  which  may  be  generated  by 
revolving  a  parabola  about  its  axis. 

IV.  An  hyperboloid  of  revolution  of  two  nappes,  which  may  be 
generated  by  revolving  an  hyperbola  about  its  transverse  axis. 


106.  These  surfaces  of  revolution  are  usually  represented  by 
taking  the  horizontal  plane  perpendicular  to  the  axis,  and  then 
drawing  the  intersection  of  the  surface  with  the  horizontal  plane; 
or  the  horizontal  projection  of  the  greatest  horizontal  circle  of  the 
surface  for  the  horizontal  projection,  and  then  projecting  on  the 
vertical  plane  the  meridian  line  which  is  parallel  to  that  plane, 
for  the  vertical  projection. 


10*7.  To  assume  a  point  on  a  surface  thus  represented,  we  first 
assume  either  projection,  as  the  horizontal,  and  erect  a  perpen- 
dicular to  the  horizontal  plane,  as  in  Art.  (76).  Through  this 
perpendicular  pass  a  meridian  plane ;  it  will  cut  from  the  surface 
a  meridian  line,  which  will  intersect  the  perpendicular  in  Vhe  re- 
quired point  or  points. 

Construction.  Let  the  surface  be  a  prolate  spheroid,  Fig.  50, 
and  let  c  be  the  horizontal,  and  c'd'  the  vertical  projection  of  the 
axis,  mon  the  horizontal  projection  of  its  largest  circle,  and  d'm'c'n' 
the  vertical  projection  of  the  meridian  curve  parallel  to  the  ver- 
tical plane,  and  let  p  be  the  assumed  horizontal  projection  ;  p  will 
be  the  horizontal,  and  s'p'  the  vertical  projection  of  the  perpen- 
dicular to  H :  cp  will  be  the  horizontal  trace  of  the  auxiliary  me- 
ridian plane.  If  this  plane  be  now  revolved  about  the  axis  until 
it  becomes  parallel  to  the  vertical  plane,  p  will  describe  the  arc 
pr,  and  r  will  be  the  horizontal,  and  r'r"  the  vertical  projection 
of  the  revolved  position  of  the  perpendicular.  The  meridian 
curve,  in  its  revolved  position,  will  be  vertically  projected  into  it? 


DESCRIPTIVE   GEOMETRY.  CI 

equal,  d'mc'n',  Art.  (61),  and  r'  and  r"  will  be.  the  vertical  pro- 
jections of  the  two  points  of  intersection  in  their  revolved  position. 
When  the  meridian  plane  is  revolved  to  its  primitive  position, 
these  points  will  describe  the  arcs  of  horizontal  circles,  projected 
on  II  in  rp,  and  on  V  in  r'p  and  r"p",  and  p'  and  p"  will  be  the 
vertical  projections  of  the  required  points  of  the  surface. 


TANGENT     PLANES    AND     TANGENT     SURFACES.        NORMAL     LINES 
AND     NORMAL     PLANES. 

108.  A  plane  is  tangent  to  a.  surface  when  it  has  at  least  one 
point  in  common  with  the  surface,  through  which,  if  any  inter- 
secting plane  he  passed,  the  right  line  cut  from  the  plane  will  be 
tangent  to  the  line  cut  from  the  surface  at  the  point.  This  point 
is  the  point  of  contact. 

It  follows  from  this  definition,  that  the  tangent  plane  is  the 
locus  of  or  place  in  which  are  to  be  found,  all  right  lines  tangent 
to  lines  of  the  surface  at  the  point  of  contact ;  and  since  any  two 
of  these  right  lines  are  sufficient  to  determine  the  tangent  plane, 
we  have  the  following  general  rule  for  passing  a  plane  tangent  to 
any  surface  at  a  given  point :  Draw  any  two  lines  of  the  surface 
intersecting  at  the  point.  Tangent  to  each  of  these,  at  the  same 
point,  draw  a  right  line.  The  plane  of  these  two  tangents  will  be 
the  required  plane. 


109.  A  right  line  r*  normal  to  a  surface,  at  a  point,  when  it  is 
perpendicular  to  the  tangent  plane  at  that  point. 

A  plane  is  normal  to  a  surface,  when  it  is  perpendicular  to  the 
iangent  plane  at  the  point  of  contact.  Since  any  plane  passed 
through  a  normal  line  will  be  perpendicular  to  the  tangent  plane 
there  may  be  an  infinite  number  of  normal  planes  to  a  surface  a 
a  point,  while  there  can  be  but  one  normal  line. 


62  DESCRIPTIVE   GEOMETRY. 

110.  The  tangent  plane  at  any  point  of  a  surface  with  recti- 
linear elements,  must  contain  the  rectilinear  elements  that  pass 
through  the  point  of  contact.  For  the  tangent  to  each  rectilinear 
element  is  the  element  itself,  Art.  (64),  and  this  tangent  must  lie 
in  the  tangent  plane,  Art.  (108). 


111.  A  plane  which  contains  a  rectilinear  element  and  its  con- 
secutive one,  of  a  single  curved  surface,  will  be  tangent  to  the  sur- 
face at  every  point  of  this  element,  or  all  along  the  element.  For 
if  through  any  point  of  the  element  any  intersecting  plane  be 
passed,  it  will  intersect  the  consecutive  element  in  a  point  con- 
secutive with  the  first  point.  The  right  line  joining  these  two 
points  will  lie  in  the  given  plane,  and  be  tangent  to  the  line  cut 
from  the  surface,  Art.  (64).  Hence  the  given  plane  will  be  tan- 
gent at  the  assumed  point. 

This  element  is  the  element  of  contact. 

Conversely,  if  a  plane  be  tangent  to  a  single  curved  surface,  it 
must,  in  general,  contain  two  consecutive  rectilinear  elements.  For 
if  through  any  point  of  the  element  contained  in  the  tangent 
plane,  Art.  (HO),  we  pass  an  intersecting  plane,  it  will  cut  from 
the  surface  a.  line,  and  from  the  plane  a  right  line,  which  will 
have  two  consecutive  points  in  common,  Art.  (108).  Through 
the  point  consecutive  with  the  assumed  point,  draw  the  consec- 
utive element  to  the  first  element;  it  must  lie  in  the  plane  of 
the  second  point  and  first  element,  Art.  (*70),  that  is,  in  the  tan- 
gent plane. 


112.  It  follows  from  these  principles,  that  if  a  plane  be  tangent 
to  a  single  curved  surface,  and  the  element  of  contact  be  inter- 
sected by  any  other  plane,  the  right  line  cut  from  the  tangent 
plane  will  be  tangent  to  the  line  cut  from  the  surface.  Hence,  if 
the  base  of  the  surface  be  in  the  horizontal  plane,  the  horizontal 
trace  of  the  tangent  plane  must  be  tangent  to  this  base,  at  the  point 


DESCRIPTIVE   GEOMETRY.  63 

in  which  the  clement  of  contact  pierces  the  horizontal  plane;  and 
the  same  principle  is  true  if  the  base  lie  in  the  vertical  plane. 


113.  A  plane  tangent  to  a  warped  surface,  although  it  contains 
the  rectilinear  element  passing  through  the  point  of  contact,  can- 
not contain  its  consecutive  element,  and  therefore  can,  in  general, 
be  tangent  at  no  other  point  of  the  element. 


114.  If  a  plane  contain  a  rectilinear  element  of  a  warped  sur- 
face, and  be  not  parallel  to  the  other  elements,  it  will  be  tangent 
to  the  surface  at  some  point  of  this  element.  For  this  plane  will 
intersect  each  of  the  other  rectilinear  elements  in  a  point;  and 
these  points  being  joined,  will  form  a  line  which  will  intersect  the 
given  element.  If  at  the  point  of  intersection  a  tangent  be  drawn 
to  this  line,  it  will  lie  in  the  tangent  plane,  Art.  (108).  The 
given  element  being  its  own  tangent,  Art.  (64),  also  lies  in  the 
tangent  plane.  The  plane  of  these  two  tangents,  that  is,  the  in- 
tersecting plane,  is  therefore  tangent  to  the  surface  at  this  point, 
Art.  (108).  Thus,  Fig.  51,  if  a  is  an  element,  b,  c,  &c,  b\  c\  &c, 
the  consecutive  elements  on  each  side  of  a,  the  plane  through  a 
will  cut  these  elements  in  the  curve  cbab'c',  which  crosses  the 
element  a  at  the  point  a.  The  tangent  mn  at  this  point,  and  the 
element  a,  determine  the  tangent  plane.  It  is  thus  seen,  that,  in 
general,  a  tangent  plane  to  a  warped  surface  is  also  an  intersecting 
plane. 

If  the  intersecting  plane  be  parallel  to  the  rectilinear  elements, 
there  will  be  no  curve  of  intersection  formed  as  above,  and  the 
plane  will  not  be  tangent. 


115.  A  plane  tangent  to  a  surface  of  revolution,  is  perpen- 
dicular to  the  meridian  plane  passing  through  the  point  of  con- 
tact.    For  this  tangent  plane  contains  the  tangent  to  the  circle  of 


64  DESCRIPTIVE    GEOMETRY. 

the  surface  at  this  point,  Art.  (108),  and  this  tangent  is  perpen- 
dicular to  the  radius  of  this  circle,  and  also  to  a  line  drawn 
through  the  point  parallel  to  the  axis,  since  it  lies  in  a  plane  per- 
pendicular to  the  axis ;  and  therefore  it  is  perpendicular  to  the 
plane  of  these  two  lines,  which  is  the  meridian  plane. 


1 


116.  Two  curved  surfaces  are  tangent  to  each  other  when  they 
have  at  least  one  point  in  common,  through  which  if  any  inter- 
secting plane  be  passed,  the  lines  cut  from  the  surfaces  will  be 
tangent  to  each  other  at  this  point.  This  will  evidently  be  the 
case  when  they  have  a  common  tangent  plane  at  this  point. 


117.  If  two  single  curved  surfaces  are  tangent  to  each  other,  at 
a  point  of  a  common  element,  tkey  will  be  tangent  all  along  this 
element.  For  the  common  tangent  plane  will  contain  this  element, 
and  be  tangent  to  each  surface  at  every  point  of  the  element, 
Art.  (111). 

This  principle  is  not  true  with  warped  surfaces. 


118.  But  if  two  warped  surfaces,  having  two  directrices,  have  a 
common  plane  directer,  a  common  rectilinear  element,  and  two  com- 
mon tangent  planes,  the  points  of  contact  being  on  the  common 
element,  they  will  be  tangent  all  along  this  element.  For  if 
through  each  of  the  points  of  contact  any  intersecting  plane  be 
passed,  it  will  intersect  the  surfaces  in  two  lines,  which  will  have, 
besides  the  given  point  of  contact,  a  second  consecutive  point  in 
common,  Art.  (108).  If,  now,  the  common  element  be  moved 
upon  the  lines  cut  from  either  surface,  as  directrices,  and  parallel 
to  the  common  plane  directer,  into  its  consecutive  position,  con- 
taining these  second  consecutive  points,  it  will  evidently  lie  in 
both  surfaces,  and  the  two  surfaces  will  thus  contain  two  consecu- 
tive rectilinear  elements.     If,  now,  any  plane  be  passed,  intersect- 


DESCRIPTIVE    GEOMETRY.  65 

ing  these  elements,  two  lines  will  be  cut  from  the  surfaces,  having 
two  consecutive  points  in  common,  and  therefore  tangent  to  each 
other;  hence  the  surfaces  will  be  tangent  all  along  the  common 
element,  Art.  (116). 


119.  Also,  if  two  warped  surfaces,  having  three  directrices,  have 
to  common  element  and  three  common  tangent  planes,  the  points  of 
contact  being  on  this  element,  they  will  be  tangent  all  along  this 
element.  For  if  through  each  point  of  contact  any  intersecting 
plane  be  passed,  it  will  intersect  the  surfaces  in  two  lines,  which, 
besides  the  given  point  of  contact,  will  have  a  second  consecutive 
point  in  common.  If  the  common  element  be  moved  upon  the 
three  lines  cut  from  either  surface,  as  directrices,  to  its  consecu- 
tive position,  so  as  to  contain  the  second  consecutive  points,  it 
will  evidently  lie  in  both  surfaces;  hence  the  two  surfaces  contain 
two  consecutive  rectilinear  elements,  and  will  be  tangent  all  along 
the  common  element. 


120.  It  follows  from  the  principle  in  Art.  (114),  that  the  pro- 
jecting plane  of  every  rectilinear  element  of  a  warped  surface  is 
tangent  to  the  surface  at  a  point.  If  through  these  points  of 
tangency,  projecting  lines  be  drawn,  they  will  form  a  cylinder 
tangent  to  the  warped  surface,  which  may  be  regarded  as  the 
projecting  cylinder  of  the  surface ;  and  the  traces  of  these  planes, 
or  the  projections  of  the  elements,  will  all  be  tangent  to  the  base 
of  this  cylinder.  This  is  seen  in  the  horizontal  projection  of  the 
hyperboloid  of  revolution  of  one  nappe,  Fig.  49  ;  also  in  both  pro- 
jections of  the  hyperbolic  paraboloid,  Fig.  45. 


121.  If  two  surfaces  of  revolution,  having  a  common  axis,  are 
tangent  to  each  other,  they  will  be  tangent  at  every  point  of  a 
circumference  of  a   circle,   perpendicular    to    the    axis.      For    if 


66  DESCRIPTIVE   GEOMETRY. 

through  the  point  of  contact  a  meridian  plane  be  passed,  it  will 
cut  from  the  surfaces  meridian  lines,  which  will  be  tangent  at  this 
point,  Art.  (108).  If  these  lines  be  revolved  about  the  common 
axis,  each  will  generate  the  surface  to  which  it  belongs,  while 
the  point  of  tangency  will  generate  a  circumference  common  to 
the  two  surfaces,  which  is  their  line  of  contact. 


SOLUTION    OF    PROBLEMS    RELATING    TO    TANGENT  PLANES  TO  SINGLE 
CURVED    SURFACES. 

122.  The  solution  of  all  these  problems  depends  mainly  upon 
the  principles,  that  a  plane  tangent  to  a  single  curved  surface  is 
tangent  all  along  a  rectilinear  element,  Art.  (ill);  and  that  if 
such  surface  and  tangent  plane  be  intersected  by  any  plane,  the 
lines  of  intersection  will  be  tangent  to  each  other. 


123.  Problem  18.  To  pass  a  plane  tangent  to  a  cylinder  at  a 
given  point  on  the  surface. 

'  Let  the  cylinder  be  given  as  in  Art.  (75),  Fig.  40,  and  let  P  be 
the  point,  assumed  as  in  Art.  (76). 

Analysis.  Since  the  required  plane  must  contain  the  rectilinear 
element  through  the  given  point,  and  its  horizontal  trace  must  be 
tangent  to  the  base  at  the  point  where  this  element  pierces  the 
horizontal  plane,  Art.  (112),  we  draw  the  element,  and  at  the 
point  where  it  intersects  the  base,  a  tangent ;  this  will  be  the 
horizontal  trace.  The  vertical  trace  must  contain  the  point  where 
this  element  pierces  the  vertical  plane,  and  also  the  point  where 
the  horizontal  trace  intersects  the  ground  line.  A  right  line  join- 
ing these  two  points  will  be  the  vertical  trace.  When  this  ele- 
ment does  not  pierce  the  vertical  plane,  within  the  limits  of  the 
drawing,  we  draw  through  any  one  of  its  points  a  line  parallel  to 
the  horizontal  trace ;  it  will  be  a  line  of  the  required  plane,  and 
pierce  the  vertical  plane  in  a  point  of  the  vertical  trace. 

Construction*  Draw  the  element  PQ ;  it  pierces  H  at  q.     At 


DESCRIPTIVE   GEOMETRY.  67 

this  point  draw  qT,  tangent  to  mlo ;  it  is  the  required  horizontal 
trace.  Through  P  draw  PZ,  parallel  to  <?T;  it  pierces  V  at  z', 
and  z'T  is  the  vertical  trace. 

mm'n'  is  the  plane  tangent  to  the  surface  along  the  element 

MN. 

124.  Problem  19.  To  pass  a  plane  through  a  given  pointy 
without  the  surface,  tangent  to  a  cylinder. 

Let  the  cylinder  be  given  as  in  the  preceding  problem. 

Analysis.  Since  the  plane  must  contain  a  rectilinear  element ; 
if  we  draw  a  line  through  the  given  point,  parallel  to  the  recti- 
linear elements  of  the  cylinder,  it  must  lie  in  the  tangent  plane, 
and  the  point  in  which  it  pierces  the  horizontal  plane,  will  be 
one  point  of  the  horizontal  trace.  If  through  this  point  we  draw 
a  tangent  to  the  base,  it  will  be  the  required  horizontal  trace.  A 
line  through  the  point  of  contact,  parallel  to  the  rectilinear  ele- 
ments, will  be  the  element  of  contact;  and  the  vertical  trace 
may  be  constructed  as  in  the  preceding  problem  :  Or  a  point  of 
this  trace  may  be  obtained  by  finding  the  point  in  which  the 
auxiliary  line  pierces  the  vertical  plane. 

Two  or  more  tangent  planes  may  be  passed,  if  two  or  more 
tangents  can  be  drawn  to  the  base,  from  the  point  in  which  the 
auxiliary  line  pierces  the  horizontal  plane. 

Let  the  construction  be  made  in  accordance  with  the  above 
analysis. 

125.  Problem  20.  To  pass  a  plane,  tangent  to  a  cylinder,  and 
parallel  to  a  given  right  line. 

Let  the  cylinder  be  given  as  in  Fig.  52,  and  let  RS  be  the 
given  line. 

Analysis.  Since  the  required  plane  must  be  parallel  to  the 
rectilinear  elements  of  the  cylinder,  as  well  as  to  the  given  line ; 
if  a.  plane  be  passed  through  this  line,  parallel  to  a  rectilinear 
element,  it  will  be  parallel  to  the  required  plane,  and  its  traces 
parallel  to  the  required  traces,  Art.  (10).     Hence,  a  tangent  to 


6$  PKSCRIPTIVK    GKOMKTEY. 

the  base,  parallel  to  the  horizontal  trace  of  this  auxiliary  plane, 
will  be  the  required  horizontal  trace.  The  element  of  contact 
and  vertical  trace  may  be  found  as  in  the  preceding  problem. 

Construction.  Through  RS  pass  the  plane  sTV,  parallel  to 
MN,  as  in  Ait.  (48).  Tangent  to  mlo  and  parallel  to  sT',  draw 
</T ;  it  is  the  horizontal  trace  of  the  required  plane,  and  UY 
parallel  to  TV  is  the  vertical  trace.  Q.P  is  the  element  of  con- 
tact. The  point  z\  determined  as  in  the  preceding  problem,  will 
aid  in  verifying  the  accuracy  of  the  drawing. 

When  more  than  one  tangent  can  be  drawn  parallel  to  6-T', 
there  will  be  more  than  one  solution  to  the  problem. 


126.  Problem  21.  To  pass  a  plane  tangent  to  a  right  cylinder 
with  a  circular  base,  having  its  axis  parallel  to  the  ground  line  at  a 
given  point  on  the  surface. 

Let  cd,  Fig.  53,  be  the  horizontal,  and  e'f  the  vertical  projec- 
tion of  the  circular  base,  and  GK  the  axis;  then  ledi  will  be  the 
horizontal,  and  u'e'f'b'  the  vertical  projection  of  the  cylinder.  Let 
p  be  the  horizontal  projection  of  the  point.  To  determine  its 
vertical  projection,  at  p  erect  a  perpendicular  to  H,  Art.  (76). 
Through  this  pass  the  plane  tTt'  perpendicular  to  AB.  It  in- 
tersects the  cylinder  in  the  circumference  of  a  circle  equal  to  the 
base,  of  which  M  is  the  centre.  This  circumference  will  intersect 
the  perpendicular  in  two  points  of  the  surface.  Revolve  this 
plane  about  ^T  until  it  coincides  with  II ;  M  falls  at  m"  Art.  (17). 
With  m"  as  a  centre  and  eg  as  a  radius,  describe  the  circle  qsr  ;  it 
will  be  the  revolved  position  of  the  circle  cut  from  the  cylinder ; 
q  and  r  will  be  the  revolved  positions  of  the  required  points,  and 
p'  andjs/'  their  vertical  projections.  Let  the  plane  be  passed 
tangent  at  P. 

Analysis.  Since  the  plane  must  contain  a  rectilinear  element, 
it  will  be  parallel  to  the  ground  line,  and  its  traces,  therefore, 
parallel  to  the  ground  line,  Art.  (9).  If  through  the  given  point 
a  plane  be  passed  perpendicular  to  the  axis,  and  a  tangent  be 


DESCRIPTIVE    GEOMETRY.  -  69 

drawn  to  its  intersection  with  the  cylinder,  at  the  point,  it  will  be 
a  line  of  the  required  plane.  If  through  the  points  in  which  this 
line  pierces  the  planes  of  projection,  lines  be  drawn  parallel  to  the 
ground  line,  they  will  be  the  required  traces. 

Construction.  Through  P  pass  the  plane  tTt',  and  revolve  it 
as  above.  At  q  draw  qx  tangent  to  qrs ;  it  is  the  revolved  posi- 
tion of  the  tangent.  When  the  plane  is  revolved  to  its  primitive 
position,  this  tangent  pierces  H  at  x,  and  V  at  y',  Ty'  being 
equal  to  TV ;  and  xz  and  y'w'  are  the  required  traces. 


127.  Problem  22.  To  pass  a  plane  tangent  to  a  right  cylinder 
with  a  circular  base,  having  its  axis  parallel  to  the  ground  line, 
through  a  given  point  without  the  surface. 

Let  the  cylinder  be  given  as  in  Fig.  53,  and  let  1ST  be  the  given 
point. 

Analysis.  Since  the  required  plane  must  contain  a  rectilinear 
element,  it  must  be  parallel  to  the  ground  line.  If  through  the 
given  point  a  plane  be  passed  perpendicular  to  the  axis,  it  will 
cut  from  the  cylinder  a  circumference  equal  to  the  base ;  and  if 
through  the  point  a  tangent  be  drawn  to  this  circumference,  it 
will  be  a  line  of  the  required  plane,  and  the  traces  may  be  deter- 
mined as  in  the  preceding  problem. 

Since  two  tangents  can  be  drawn,  there  may  be  two  tangent 
planes. 

Let  the  construction  be  made  in  accordance  with  the  analysis. 

128.  Problem  23.  To  pass  a  plane  parallel  to  a  given  right 
line,  and  tangent  to  a  right  cylinder,  with  a  circular  base,  with  its 
axis  parallel  to  the  ground  line. 

Let  the  cylinder  be  given  as  in  Fig.  54,  and  let  MN  be  the 
given  line. 

Analysis.  Since  the  plane  must  be  parallel  to  the  axis,  if 
through  the  given  line  we  pass  a  plane  parallel  to  the  axis,  it  will, 
be  parallel  to  the  required  plane,  and  to  the  ground  line. 


70  -  DESCRIPTIVE  GEOMETRY. 

A  plane  perpendicular  to  the  ground  line  will  cut  from  the 
cylinder  a  circumference  ;  from  the  required  tangent  plane  a  tan- 
gent to  this  circumference,  Art.  (108);  and  from  the  parallel 
plane  a  right  line  parallel  to  the  tangent.  If,  then,  we  construct 
the  circumference,  and  draw  to  it  a  tangent  parallel  to  the  inter- 
section of  the  parallel  plane,  this  tangent  will  be  a  line  of  the  re- 
quired plane,  from  which  the  traces  may  be  found  as  in  the 
preceding  problems. 

Construction.  Through  m  and  n'  draw  the  lines  mp  and  n'q', 
parallel  to  AB.  They  will  be  the  traces  of  the  parallel  plane, 
Art.  (48).  Let  tTt'  be  the  plane  perpendicular  to  AB.  It  cuts 
from  the  cylinder  a  circle  whose  centre  is  O,  and  from  the  parallel 
plane  a  right  line  which  pierces  H  at  p  and  V  at  q\  Art.  (38). 
Revolve  this  plane  about  tT,  until  it  coincides  with  H ;  xyz  will 
be  the  revolved  position  of  the  circle,  and  sp  that  of  the  line  cut 
from  the  parallel  plane  ;'  zu  tangent  to  xyz,  and  parallel  to  sp,  will 
be  the  revolved  position  of  a  line  of  the  required  plane,  which,  in 
its  true  position,  pierces  H  at  w,  and  V  at  w\  and  uv  and  w'v'  are 
the  traces  of  the  required  plane. 

Since  another  parallel  tangent  can  be  drawn,  there  will  be  two 
solutions. 

129.  Problem  24.  To  pass  a  plane  tangent  to  a  cone,  through 
a  given  point  on  the  surface. 

Let  the  cone  be  given  as  in  Fig.  41,  and  let  P,  assumed  as  in 
Art.  (79),  be  the  given  point. 

Analysis.  The  required  plane  must  contain  the  rectilinear 
element,  passing  through  the  given  point,  Art.  (110).  If,  then, 
we  draw  this  element,  Art.  (79),  and  at  the  point  where  it 
pierces  the  horizontal  plane,  draw  a  tangent  to  the  base,  it  will 
be  the  horizontal  trace  of  the  required  plane,  and  points  of  the 
vertical  trace  may  be  found  as  in  the  similar  case  for  the  cyl- 
inder. 

Let  the  construction  be  made  in  accordance  with  the  anal- 
ysis. 


DESCRIPTIVE   GEOMETRY.  71 

130.  Problem  25.  Through  a  point  without  the  surface  of  a 
tone,  to  pass  a  plane  tangent  to  the  cone. 

Let  the  cono  be  given  as  in  Fig.  55,  and  let  P  be  the  given 
point. 

Analysis.  Since  the  required  plane  must  contain  a  rectilinear 
element,  it  will  pass  through  the  vertex;  hence,  if  we  join  the 
given  point  with  the  vertex  by  a  right  line,  it  will  be  a  line  of  the 
required  plane,  and  pierce  the  horizontal  plane  in  a  point  of  the 
required  horizontal  trace,  which  may  then  be  drawn  tangent  to 
the  base.  If  the  point  of  contact  with  the  base  be  joined  to  the 
vertex  by  a  right  line,  it  will  be  the  element  of  contact.  The 
vertical  trace  may  be  found  as  in  the  preceding  problems. 

If  more  than  one  tangent  can  be  drawn  to  the  base,  there  will 
be  more  than  one  solution. 

Construction.  Join  P  with  S,  by  the  line  PS ;  it  pierces  H 
at  u.  Draw  uq  tangent  to  mlo\  it  is  the  required  horizontal 
'  trace.  SQ  is  the  element  of  contact  which  pierces  V  at  w\  and 
z'T  is  the  vertical  trace. 

ux  will  be  the  horizontal  trace  of  a  second  tangent  plane 
through  P. 


131.  Problem  26.  To  pass  a  plane  tangent  to  a  cone,  and 
parallel  to  a  given  right  line. 

Let  the  cone  be  given  as  in  Fig.  55,  and  let  NR  be  the  given 
right  line. 

Analysis.  If  through  the  vertex  we '  draw  a  line  parallel  to 
the  given  line,  it  must  lie  in  the  required  plane,  and  pierce  the 
horizontal  plane  in  a  point  of  the  horizontal  trace.  Through  this 
point  draw  a  tangent  to  the  base,  it  will  be  the  required  horizon- 
tal trace ;  and  the  element  of  contact  and  vertical  trace  may  be 
found  as  in  the  preceding  problem. 

When  more  than  one  tangent  can  be  drawn  to  the  base,  there 
will  be  more  than  one  solution. 

If  the  parallel  line  through  the  vertex  pierces  the  horizonta 


72  DESCRIPTIVE   GEOMETRY. 

plane  within  the  base,  no  tangent  can  be  drawn,  and  the  problem 
is  impossible. 

Let  the  construction  be  made  in  accordance  with  the  analysis. 


132.  Problem  27.  To  pass  a  plane  tangent  to  a  single  curved 
surface  with  a  helical  directrix,  at  a  given  point. 

Let  the  surface  be  given  as  in  Fig.  42,  and  let  R  be  the  given 
point,  Art.  (81). 

Analysis.  The  tangent  plane  must,  in  general,  be  tangent  all 
along  the  rectilinear  element,  through  the  point  of  contact.  Its 
horizontal  trace  must  therefore  be  tangent  to  the  base,  Art.  (112). 
If  through  the  point  where  this  element  pierces  the  horizontal 
plane,  a  tangent  be  drawn  to  the  base,  it  will  be  the  required 
horizontal  trace,  and  the  vertical  trace  may  be  determined  as  in 
the  case  of  the  cylinder  or  cone. 

Construction.  The  element  RX  pierces  H  at  Z.  At  thiu 
point  draw  the  tangent  ^T ;  it  is  the  horizontal  trace.  Trf'  is  the 
vertical  trace. 

To  pass  a  plane  through  a  given  point  without  this  surface, 
tangent  to  it,  we  pass  a  plane  through  the  point  parallel  to  the 
base,  and  draw  a  tangent  to  the  curve  of  intersection,  Art.  (81), 
through  the  point.  This  tangent,  with  the  element  of  the  surface 
through  its  point  of  contact,  will  determine  the  tangent  plane. 

133.  Problem.  28.  To  pass  a  plane  tangent  to  a  single  curved 
surface  with  a  helical  directrix,  and  parallel  to  a  given  right  line. 

Let  the  surface  be  given  as  in  Fig.  42,  and  let  MN  be  the 
given  line. 

Analysis.  If  with  any  point  of  the  right  line,  as  a  vertex,  we 
construct  a  cone,  whose  elements  make  the  same  angle  with  the 
horizontal  plane  as  the  elements  of  the  surface ;  and  pass  a  plane 
through  the  line  tangent  to  this  cone,  it  will  be  parallel  to  the 
required  plane.  The  traces  of  the  required  plane  may  then  be 
constructed  as  in  Art.  (125). 


DESCRIPTIVE   GEOMETRY.  73 

Construction.  Take  n'  as  the  vertex  of  the  auxiliary  cone, 
and  draw  no",  making  with  AB  an  angle  equal  to  oVT;  o"cs" 
will  be  the  base  of  the  cone  in  II.  Through  m  draw  mc  tangent 
to  o"cs".  It  will  be  the  horizontal  trace  of  the  parallel  plane, 
and  t"T,  parallel  to  it,  and  tangent  to  uvw,  is  the  required  hori- 
zontal trace.     Let  the  pupil  construct  the  vertical  trace. 

134.  It  is  a  remarkable  property  of  this  surface,  with  a  helical 
directrix,  that  any  plane  passing  through  a  rectilinear  element,  is 
tangent  to  the  surface,  at  the  point  where  the  element  intersects 
the  directrix.  For  this  plane  will  intersect  the  other  elements, 
thus  forming  a  curve,  Art.  (114).  This  curve  will  intersect  the 
given  element  at  the  point  where  the  element  touches  the  direc- 
trix. The  tangent  to  the  curve  at  this  point,  and  the  element, 
both  lie  in  the  given  plane ;  it  is  therefore  tangent  to»the  surface 
at  the  point,  Art.  (114).  This  plane  does  not,  in  general,  contain 
the  consecutive  element,  and  is  therefore  not  necessarily  tangent 
all  along  the  element. 

The  projecting  planes  of  all  the  elements  are  tangent  to  the 
surface,,  and  the  cylinder  formed  by  the  projecting  lines  of  the 
points  of  contact  is  therefore  tangent  to  the  surface,  Art.  (120). 
Its  base  is  the  circle  pxq. 

It  should  be  observed,  also,  that  at  any  point  on  the  helical  di- 
rectrix, an  infinite  number  of  planes  can  be  passed  tangent  to  the 
surface,  as  at  the  vertex  of  a  cone.  Only  one  of  these  planes  will 
contain  two  consecutive  rectilinear  elements. 


135.  By  an  examination  of  the  preceding  problems,  it  will  be 
seen  that,  with  two  remarkable  exceptions,  only  one  tangent  plane 
can  be  drawn  to  a  single  curved  surface  at  a  given  point. 

That  the  number  which  can  be  drawn  through  a  given  point 
without  the  surface,  and  tangent  along  an  element,  will  be  limited. 

That  the  number  which  can  be  drawn  parallel  to  a  given  right 
ine,  and  tangent  along  an  element,  is  also  limited. 


74:  DESCRIPTIVE    GEOMETRY. 

That,  in  general,  a  plane  cannot  be  passed  through  a  gm.n 
right  line  and  tangent  to  a  single  curved  surface.  If,  however, 
the  given  line  lies  on  the  convex  side  of  the  surface,  and  is  par- 
allel-to  the  rectilinear  elements  of  a  cylinder,  or  passes  through  the 
vertex  of  a  cone,  or  is  tangent  to  a  line  of  the  surface,  the  prob- 
lem is  possible. 


PROBLEMS    RELATING    TO    TANGENT     PLANES    TO    WARPED    SURFACES. 

136.  Since  a  tangent  plane  to  a  warped  surface  must  contain 
the  rectilinear  element  passing  through  the  point  of  contact,  Art 
(110),  Ave  can  at  once  determine  one  line  of  the  tangent  plane. 
A  second  line  may  then  be  determined,  in  accordance  with  the 
rule  in  Art.  (108). 

When  the  surface  has  two  different  generations  by  right  lines, 
the  plane  of  the  two  rectilinear  elements  passing  through  th.e 
given  point  will  be  the  required  plane. 


137.  Problem  29.  To  pass  a  plane  tangent  to  a  hyperbolic 
paraboloid,  at  a  given  point  of  the  surface. 

Let  MN  and  PQ,  Fig.  56,  be  the  directrices,  and  MP  and  NQ 
any  two  elements  of  the  surface,  and  let  Q,  assumed  as  in  Art. 
(88),  be  the  given  point. 

Analysis.  Since  through  the  given  point  a  rectilinear  element 
of  each  generation  can  be  drawn,  Art.  (89),  we  have  simply  to 
construct  these  two  elements,  and  pass  a  plane  through  them, 
Art.  (136). 

Construction.  Through  0  draw  OE  and  OF,  parallel  respect- 
ively to  NQ  and  MP.  These  will  determine  a  plane  parallel  to 
the  plane  directer  of  the  first  generation,  Art.  (86).  This  plane 
cuts  the  directrix  PQ  in  the  point  U,  Art.  (41).  Join  U  with  O 
and  we  have  an  element  of  the  first  generation.  Art.  (83).  Let 
NQ  and  MI*  be  taken  as  directrices  of  the  second  generation. 


DESCRIPTIVE   GEOMETRY.  75 

Through  O  draw  OC  and  OD  parallel  respectively  to  MN  and 
PQ.  They  will  determine  a  plane  parallel  to  the  plane  directer 
of  the  second  generation,  Art.  (86).  This  plane  cuts  MP  in  W 
and  OW  will  be  an  element  of  the  second  generation,  and  the 
tangent  plane  is  determined,  as  in  Art.  (32). 


138.  Problem  30.     To  pass  a  plane  tangent  to  an  hyperboloi 
of  revolution  of  one  nappe,  at  a  given  point  of  the  surface. 

Let  the  surface  be  given  as  in  Fig.  57,  and  let  0  be  the  given 
point. 

Analysis.  Same  as  in  the  preceding  problem. 

Construction.  OZ  is  the  element  of  the  first  generation  parsing 
through  O,  Art.  (100).  It  pierces  H  at  r.  Draw  OS,  Art.  (101). 
It  is  an  element  of  the  second  generation  passing  through  0. 
This  element  pierces  H  at  w,  and  wr  is  the  horizontal  trace  of  the 
required  plane,  and  Ttf  is  its  vertical  trace. 

Since  the  meridian  plane  through  O  must  be  perpendicular  to 
the  tangent  plane,  Art.  (115),  its  trace  m  must  be  perpendicular 
to  wr. 


139.  Problem  31.  To  pass  a  plane  tangent  to  a  helicoid.  at  a 
point  of  the  surface. 

Let  the  surface  be  given  as  in  Fig.  48,  and  let  M  be  the  given 
point. 

Analysis.  The  tangent  plane  must  contain  the  rectilinear  el- 
ement* passing  through  the  given  point,  and  also  the  tangent  to 
the  helix  at  this  point,  Art.  (108).  The  plane  of  these  two  lines 
will  then  be  the  required  plane. 

Construction.  MX  is  the  rectilinear  element  through  M.  I 
pierces  II  at  u ;  cmd  is  the  horizontal  projection  of  the  heli> 
through  M.  Draw  the  tangent  to  this  helix  at  M,  as  in  Art 
(09).  mz  will  be  its  horizontal  projection  ;  Z  the  point  in  which 
it  pierces  the  horizontal   plane  through  C,  and  m'z    its  vertical 


76  DESCRIPTIVE    GEOMETRY. 

projection.     This  tangent  pierces  H  at  h\   hence  Tcu  is  the  hori- 
zontal trace  of  the  required  plane,  and  t'T  is  the  vertical  trace. 

Since  a  tangent  plane  to  the  surface  contains  a  rectilinear  el- 
ement, it  is  evident  that  it  cannot  make  a  less  angle  with  the  hor- 
izontal plane  than  the  elements;  nor  a  greater  angle  than  90°, 
the  angle  made  by  the  projecting  plane  of  any  rectilinear  element, 
which  is  tangent  at  the  point  where  the  element  intersects  the 


140.  Problem  32.  To  pass  a  plane  tangent  to  a  helicoid,  and 
perpendicular  to  a  given  right  line. 

Analysis.  If,  with  any  point  of  the  axis  as  a  vertex,  we  con- 
struct a  cone  whose  rectilinear  elements  shall  make  with  the  hor- 
izontal plane  the  same  angle  as  that  made  by  the  rectilinear  el- 
ements of  the  given  surface,  and  through  the  vertex  of  this  cone 
pass  a  plane  perpendicular  to  the  given  line,  Art.  (46),  it  will,  if 
the  problem  be  possible,  cut  from  the  cone  two  elements,  each  of 
which  will  be  parallel  to^,  rectilinear  element  of  the  helicoid,  and 
have  the  same  horizontal  projection.  If  through  eiihcv  of  these 
elements  a  plane  be  passed  parallel  to  the  auxiliary  pkne,  it  will 
be  tangent  to  the  surface,  Art.  (114),  and  perpc.dku'br  to  the 
given  line. 

Let  the  problem  be  constructed  in  accordance  with  the  analysis. 


141.  An  auxiliary  surface  may  sometimes  be  used  to  advantage 
in  passing  a  plane  tangent  to  a  warped  surface  at  a  given 'point 
Thus,  let  MN  and  PQ,  Fig.  58,  be  the  two  directrices  of  a  warped 
surface  having  V  for  its  plane  directer,  and  let  O  be  the  given 
point.  At  the  points  X  and  Y,  in  which  the  rectilinear  element 
through  O  intersects  the  directrices,  draw  a  tangent  to  each  direc- 
trix, as  XZ  and  YU.  On  these  tangents  as  directrices,  mov^  XY 
parallel  to  V.  It  will  generate  an  hyperbolic  paraboloid,  A\i 
(86),  having,  with  the  given  surface,  the  common  element  X"S 


DESCRIPTIVE    GEOMETRY.  77 

and  a  common  tangent  plane  at  each  of  the  points  X  and  Y,  since 
the  plane  of  the  two  lines  XY  and  XZ,  and  also  that  of  XY  and 
YU,  is  tangent  to  both  surfaces,  Art.  (108).  The  two  surfaces  are 
therefore  tangent  all  along  the  common  element,  XY,  Art.  (118) 
If,  then,  at  O,  we  pass  a  plane  tangent  to  the  hyperbolic  parabo- 
loid, it  will  be  also  tangent  to  the  given  surface  at  the  same  point. 
If  the  given  surface  have  three  curvilinear  directrices,  a  tangent 
may  be  drawn  to  each,  at  the  point  in  which  the  rectilinear 
element  through  the  given  point  intersects  it ;  and  then  this  el- 
ement may  be  moved  on  these  three  tangents  as  directrices,  gen- 
erating a  hyperboloid  of  one  nappe,  Art.  (91),  which  will  be 
tangent  to  the  given  surface  all  along  a  common  element,  Art. 
(119).  A  plane  tangent  to  this  auxiliary  surface  at  the  given, 
point,  will  also  be  tangent  to  the  given  surface. 


142.  An  infinite  number  of  planes  may,  in  general,  be  passed 
through  a  point  without  a  warped  surface,  and  tangent  to  it.  Fot 
if,  through  the  point,  a  system  of  planes  be  passed  intersecting 
the  surface,  tangents  may  be  drawn  from  the  point  to  the  curves 
of  intersection,  and  these  will  form  the  surface  of  a  cone  tangent 
to  the  warped  surface.  Any  plane  tangent  to  this  cone  will  be 
tangent  to  the  warped  surface,  and  pass  through  the  point. 

Also,  an  infinite  number  of  planes  may,  in  general,  be  passed 
tangent  to  a  warped  surface,  and  parallel  to  a  right  line.  For  if 
the  surface  be  intersected  by  a  system  of  planes  parallel  to  the 
line,  and  tangents  parallel  to  the  line  be  drawn  to  the  sections^ 
they  will  form  the  surface  of  a  cylinder,  tangent  to  the  warped 
surface.  Any  plane  tangent  to  this  cylinder  will  be  tangent  to 
the  warped  surface,  and  parallel  to  the  given  line. 


143.    To  pass  a  plane  through  a  given  right  line,  and  tangen 
to  a  warped  surface,  it  is  only  necessary  to  produce  the  line  unti 


78  DKSCRIPTIVE    GEOMETRY. 

it  pierces  the  surface,  and  through  the  point  thus  determined 
draw  the  rectilinear  element  of  the  surface.  This,  with  the  given 
line,  will  determine  a  plane  tangent  to  the  surface  at  some  point 
of  the  element,  Art.  (114). 

If  there  be  two  rectilinear  elements  passing  through  this  point, 
each  will  give  a  tangent  plane ;  and  the  number  of  tangent  planes 
will  depend  upon  the  number  of  points  in  which  the  line  pierces 
the  surface.  If  the  given  line  be  parallel  to  a  rectilinear  element, 
it  pierces  the  surface  at  an  infinite  distance,  and  the  tangent  plane, 
will  be  determined  by  the  two  parallel  lines. 


PROBLEMS     RELATING    TO.     TANGENT    PLANES     TO     DOUBLE     CURVED 
SURFACES. 

144.  These  problems  are,  in  general,  solved  either  by  a  direct 
application  of  the  rule  in  Art.  (108),  taking  care  to  intersect  the 
surface  by  planes,  so  as  to  obtain  the  two  simplest  curves  of  the 
surface  intersecting  at  the  point  of  contact,  or  by  means  of  more 
simple  auxiliary  surfaces  tangent  to  the  given  surface. 


145.  Problem  33.  To  pass  a  plane  tangent  to  a  sphere,  at  a 
given  point. 

Let  C,  Fig.  59,  be  the  centre  of  the  sphere ;  prq  its  horizontal, 
and  p'x'q'  its  vertical  projection.  Let  M  be  the  point  assumed, 
as  in  Art.  (107). 

Analysis.  Since  the  radius  of  the  sphere,  drawn  to  the  point 
of  contact,  is  perpendicular  to  the  tangent  to  any  great  circle  at 
this  point,  and  since  these  tangents  all  lie  in  the  tangent  plane, 
Art.  (108),  this  radius  must  be  perpendicular  to  the  tangent  plane. 
We  have  then  simply  to  pass  a  plane  perpendicular  to  this  radius 
at  the  given  point,  and  it  will  be  the  required  plane. 

Construction.  This  may  be  made  directly,  as  in  Art.  (46),  or 


DESCRIPTIVE   GEOMETRY.  79 

otherwise,  thus :  Through  M  and  C  pass  a  plane  perpendicular 
to  II ;  cm  will  be  its  horizontal  trace.  Revolve  this  plane  about 
cm  as  an  axis,  until  it  coincides  with  H ;  c"m"  will  be  the  re- 
volved position  of  the  radius,  Art.  (28).  At  m"  draw  m"t  per- 
pendicular to  c"m".  It  is  the  revolved  position  of  a  line  of  the 
required  plane.  It  pierces  H  at  t,  and  since  the  horizontal  trace 
must  be  perpendicular  to  cm,  Art.  (43),  tT  is  the  required  hori- 
zontal trace.  Through  II  draw  MN  parallel  to  tT.  It  pierces  V 
at  n  ;   and  Tn\  perpendicular  to  c'm',  is  the  vertical  trace. 


146.  Problem  34.  To  pass  a  plane  tangent  to  an  ellipsoid  of 
revolution,  at  a  given  point. 

Let  the  surface  be  given  as  in  Art.  (107),  Fig.  50,  and  let  P  be 
the  point. 

Analysis.  If,  at  the  given  point,  we  draw  a  tangent  to  the 
meridian  curve  of  the  surface,  and  a  second  tangent  to  the  circle 
of  the  surface  at  this  point,  the  plane  of  these  two  lines  will  be 
the  required  plane,  Art.  (108). 

Construction.  Through  P  pass  a  meridian  plane  ;  cp  will  be 
its  horizontal  trace.  Revolve  this  plane  about  the  axis  of  the  sur- 
face until  it  is  parallel  to  Y.  R  will  be  the  revolved  position  of 
the  point  of  contact.  At  r'  draw  r'x'  tangent  to  c'nd'm' .  It 
will  be  the  vertical  projection  of  the  revolved  position  of  a  tangent 
to  the  meridian  curve  at  R.  When  the  plane  is  revolved  to  its 
primitive  position,  the  point,  of  which  y'  is  the  vertical  projection, 
remains  fixed,  and  y'p'  will  be  the  vertical  projection  of  the  tan 
gent,  and  cp  its  horizontal  projection.  It  pierces  H  at  ar,  one 
point  of  the  hori  jontal  trace  of  the  required  plane,  pr  is  the 
horizontal,  and  pr'  the  vertical  projection  of  an  arc  of  t!ie  circle 
of  the  surface  containing  P.  At  p  draw  pa  perpendicular  to  cp. 
It  will  be  the  horizontal  projection  of  the  tangent  to  the  circle 
at  P,  and  p'u'  is  its  vertical  projection.  This  pierces  V  at  u1 ',  a 
point  of  the  vertical  trace.  Through  x  draw  xT,  parallel  to  pu, 
and  through  T  draw  TV ;  xTu'  will  be  the  required  plane. 


80  DESCRIPTIVE    GKOMETKT. 

147.  Second  method  for  the  same  problem. 

Analysis.  If  the  tangent  to  the  meridian  curve  at  P  revolve 
about  the  axis  of  the  surface,  it  will  generate  aright  cone  tangent 
to  the  surface  in  a  circumference  containing  the  given  point,  Art. 
(121).  If,  at  this  point,  a  tangent  plane  be  drawn  to  the  cone,  it 
will  be  tangent  also  to  the  surface. 

Construction.  Draw  the  tangent  at  P,  as  in  the  preceding 
article.  It  pierces  H  at  a*,  and  this  point,  during  the  revolution, 
describes  the  base  of  the  cone  whose  vertex  is  at  (eg').  The  ele- 
ment of  the  cone  through  P  pierces  H  at  x,  and  xT?  is  the  re- 
quired plane. 

148.  By  the  same  methods  a  tangent  plane  may  be  passed  to 
any  surface  of  revolution  at  a  given  point. 

Since  the  tangent  plane  and  horizontal  plane  are  both  perpen- 
dicular to  the  meridian  plane  through  the  point  of  contact,  their 
intersection,  which  is  the  horizontal  trace,  will  be  perpendicular 
to  the  meridian  plane  and  to  its  horizontal  trace. 

While  at  a  given  point  on  a  double  curved  surface  only  one 
tangent  plane  can  be  passed,  it  may  be  proved  as  in  Art.  (142), 
that  from  a  point  without  the  surface,  an  infinite  number  of  such 
planes  can  be  passed. 

149.  Problem  35.  To  pass  a  plane  through  a  given  right  line 
and  tangent  to  a  sphere. 

Let  the  ground  line  pass  through  the  centre  of  the  sphere,  and 
let  C,  Fig.  60,  be  the  centre,  and  def  the  circle,  cut  from  the 
sphere  by  the  horizontal  plane,  and  let  MN  be  the  given  line.. 

Analysis.  If  we  take  any  point  of  the  given  line  as  the  vertex 
of  a  right  cone,  tangent  to  the  sphere,  and  pass  a  plane  through 
the  line  tangent  to  this  cone,  it  will  be  tangent  also  to  the  sphere 

Construction.  Take  m  as  the  vertex  of  the  auxiliary  cone,  and 
draw  the  tangents  md  and  me,  and  the  right  line  wiC.  The  tan- 
gents will  be  the  two  rectilinear  elements  of  the  cone  in  the  hori- 


DESCRIPTIVE    GEOMETRY.  81 

zontal  plane,  and  mC  will  be  its  axis.  Since  the  line  of  contact 
of  the  two  surfaces  is  a  circumference,  whose  plane  is  perpendic- 
ular to  mC,  Art.  (121),  this  plane  will  be  perpendicular  to  II,  and 
de  will  be  its  horizontal  trace.  This  circumference  may  be  re- 
garded as  the  base  of  the  cone;  and,  if  we  find  the  point  in 
which  its  plane  is  pierced  by  MN,  and  draw  from  this  point  a 
tangent  to  the  base,  it  will  be  a  line  of  the  required  plane,  Art. 
(112).  0  is  this  point,  Art.  (15).  If  the  plane  of  the  base  be 
revolved  about  de  as  an  axis,  until  it  coincides  with  II,  O  will 
fall  at  o",  o"o  being  equal  to  o'q,  Art.  (1*7).  The  circle  of  con- 
tact will  take  the  position  dp"e,  de  being  its  diameter.  Draw 
o"p"\  it  will  be  the  revolved  position  of  the  tangent.  It  pierces 
H  at  t,  one  point  of  the  horizontal  trace ;  MN,  pierces  H  at  m; 
hence,  tm  is  the  required  horizontal  trace.  MN  pierces  V  at  n', 
a  point  of  the  vertical  trace.  A  second  point,  t\  may  be  deter- 
mined as  in  Art.  (123),  and  n't'  is  the  vertical  trace.  When  the 
plane  of  the  circle  of  contact  is  in  its  true  position,  p"  is  hori- 
zontally projected  at  p,  and  vertically  at  p',  p'r  being  equal  to 
pp"  ;  hence,  CPwill  be  the  radius  passing  through  P.  Since  the 
tangent  plane  must  be  perpendicular  to  this  radius,  Cp  and  Qp' 
must  be  respectively  perpendicular  to  tm  and  tfnf. 

Since  a  second  tangent  can  be  drawn  from  O  to  the  base  of 
the  cone,  another  tangent  plane  may  be  constructed. 


150.    Second  method  for  the  same  problem. 

Let  the  sphere  and  the  right  line  MN  be  given  as  in  Fig.  61. 

Analysis.  If  any  two  points  of  the  given  line  be  taken,  each 
as  the  vertex  of  a  cone  tangent  to  the  sphere,  each  cone  will 
be  tangent  in  the  circumference  of  a  circle,  and  the  planes  of 
these  circles  will  intersect  in  a  right  line,  which  will  pierce  the 
surface  of  the  sphere  at  the  intersection  of  the  circumferences, 
and  these  points  will  be  common  to  the  three  surfaces.  If, 
through  either  of  these  points  and  the  given  line,  we  pass  a 
plane,  it  will  be  tangent  to  both  cones  and  to  the  sphere. 


82  DESCRIPTIVE   GEOMETRY. 

Construction*  Take  the  points  m  and  n'  as  the  vertices  of  the 
two  auxiliary  cones ;  de  is  the  horizontal  projection  of  the  circle 
of  contact  of  the  first  cone  with  the  sphere,  and  fg  is  the  vertical 
projection  of  the  circle  of  contact  of  the  second  cone  and  sphere, 
Art.  (15).  The  planes  of  these  two  circles  intersect  in  a  right 
line,  of  which  de  is  the  horizontal,  and  fg  the  vertical  projection, 
Art.  (15) ;  and  this  line  pierces  H  at  o.  Now  revolve  the  circle 
of  which  di  is  the  diameter,  about  de  as  an  axis,  until  it  coin- 
cides with  H.  It  takes  the  position  dsex.  Any  point  of  the  line 
{de,  fg),  as  R,  falls  at  r",  and  or"  will  be  the  revolved  position  oi 
the  line  of  intersection  of  the  two  planes,  and  p"  and  q"  will 
be  the  revolved  positions  of  the  two  points  in  which  it  pierces 
the  surface  of  the  sphere.  After  the  counter-revolution,  these 
points  are  horizontally,  projected  at  p  and  q,  and  vertically  at  p' 
and  q'.  CP  is  the  radius  of  the  sphere  at  P.  The  traces  tT,  and 
t'n',  of  the  plane  tangent  at  P,  may  now  be  drawn  as  in  the  pre- 
ceding article ;  or  by  drawing  mt  perpendicular  to  Cp,  and  n't' 
perpendicular  to  Cp',  Art.  (43).  A  second  tangent  plane  at  Q 
may  be  determined  in  the  same  way. 


161.    Third  method  for  the  same  problemk 

Let  C,  Fig.  62,  be  the  centre,  and  def  the  horizontal,  and  d'h'f 
the  vertical  projection  of  the  sphere,  and  let  MN  be  the  given 
line. 

Analysis.  Conceive  the  sphere  to  be  circumscribed  by  a  cylin- 
der of  revolution,  whose  axis  is  parallel  to  the  given  line.  The 
line  of  contact  will  be  the  circumference  of  a  great  circle  perpen- 
dicular to  the  axis  and  given  line,  Art.  (121).  A  plane  through 
the  right  line  tangent  to  this  cylinder  will  be  tangent  also  to  the 
sphere.  The  plane  of  the  circle  of  contact  will  intersect  the 
given  line  in  a  point,  and  the  required  tangent  plane  in  a  right 
line  drawn  from  this  point  tangent  to  the  circle.  The  plane  of 
this  tangent  and  the  given  line  will  be  the  required  plane.  With- 
out constructing  the  cylinder,  we  have  then   simply  to  pass  a 


DESCRIPTIVE    GEOMETRY.  83 

plane  through  the  centre  of  the  sphere  perpendicular  to  the  given 
line,  and  from  the  point  in  which  it  intersects  the  line,  to  draw  a 
tangent  to  the  circle  cut  from  the  sphere  by  the  same  plane,  and 
pass  a  plane  through  this  tangent  and  the  given  line. 

Construction.  Through  C  draw  the  two  lines  CP  and  CQ,  as 
in  Art.  (46).  The  plane  of  these  two  lines  is  perpendicular  to 
MN,  and  intersects  it  in  0,  Art.  (41).  The  horizontal  trace  or 
this  plane  may  now  be  determined  as  in  Art.  (46),  and  the  plane 
revolved  about  this  trace  until  it  coincides  with  H,  and  the  re- 
volved position  of  the  point  and  circle  be  found,  and  the  tangent 
drawn. 

.  Otherwise  thus :  Revolve  this  plane  about  CP,  until  it  becomes 
parallel  to  H.  The  point  0,  in  its  revolved  position,  will  be  hori- 
zontally projected  at  r,  and  the  circle  will  be  horizontally  pro- 
jected in  def,  Art.  (62).  From  r  draw  the  two  tangents  rk  and 
rl.  They  will  be  the  horizontal  projections  of  the  revolved  po- 
sitions of  the  two  tangents  to  the  great  circle,  cut  from  the  sphere 
by  the  perpendicular  plane,  and  k  and  I  will  be  the  horizontal 
projections  of  the  revolved  positions  of  the  two  points  of  contact 
of  the  required  tangent  plane.  After  the  counter-revolution,  R 
will  be  horizontally  projected  at  o;  and  since  X  remains  fixed,  oy 
will  be  the  horizontal,  and  o'y'  the  vertical  projection  of  the  first 
tangent,  and  Y  its  point  of  contact. 

Since  the  second  tangent  does  not  intersect  PC  within  the 
limits  of  the  drawing,  draw  the  auxiliary  line  gp,  and  find  the 
true  horizontal  projection  of  the  point  which  in  revolved  position 
is  horizontally  projected  in  g,  as  in  Art.  (37).  It  will  be  at  g'\ 
and  og"  will  be  the  horizontal  projection  of  the  second  tangent, 
and  z  the  horizontal  projection  of  the  second  point  of  contact. 
S  is  the  point  in  which  this  tangent  intersects  CQ,  o's'  is  its  verti- 
cal projection,  and  Z  is  the  second  point  of  contact.  A  plane 
through  MN,  and  each  of  these  points  will  be  tangent  to  the 
sphere. 


84:  DESCRIPTIVE    GEOMETRY. 

152.  Problem  36.  To  pass  a  plane  through  a  given  right  line 
and  tangent  to  any  surface  of  revolution. 

Let  the  horizontal  plane  be  taken  perpendicular  to  the  axis,  of 
which  c,  Fig.  63,  is  the  horizontal,  and  c'd'  the  vertical  projection. 
Let  poq  be  the  intersection  of  the  surface  by  the  horizontal  plane, 
p'd'q  the  vertical  projection  of  the  meridian  curve  parallel  to  the 
vertical  plane,  and  MN  the  given  line. 

Analysis.  If  this  line  revolve  about  the  axis  of  the  surface,  it 
will  generate  a  hyperboloid  of  revolution  of  one  nappe,  Art.  (99), 
having  the  same  axis  as  the  given  surface.  If  we  now  conceive 
the  plane  to  be  passed  tangent  to  the  surface,  it  will  also  be  tan- 
gent to  the  hyperboloid  at  a  point  of  the  given  right  line,  Art. 
(114) ;  and  since  the  meridian  plane  through  the  point  of  contact 
on  each  surface  must  be  perpendicular  to  the  common  tangent 
plane,  Art.  (115),  these  meridian  planes  must  form  one  and  the 
same  plane.  This  plane  will  cut  from  the  given  surface  a  me- 
ridian curve,  from  the  hyperboloid  an  hyperbola,  Art.  (104),  and 
from  the  tangent  plane  a  right  line  tangent  to  these  curves  at  the 
required  points  of  contact,  Art.  (108).  The  plane  of  this  tangent 
and  the  given  line  will  therefore  be  the  required  plane. 

Construction.  Construct  the  hyperboloid  as  in  Art.  (99).  ex 
will  be  the  horizontal,  and  x'y'x"  the  vertical  projection  of  one 
,  branch  of  the  meridian  curve  parallel  to  V.  If  the  meridian 
plane  through  the  required  points  of  contact  be  revolved  about 
the  common  axis  until  it  becomes  parallel  to  V,  the  corresponding 
meridian  curves  will  be  projected,  one  into  the  curve  p'd'q',  and 
the  other  into  the  hyperbola  x'y'x".  Tangent  to  these  curves 
draw  x'r' ;  X  will  be  the  revolved  position  of  the  point  of  con 
tact  on  the  hyperbola,  and  11  that  on  the  meridian  curve  of  the 
given  surface.  When  the  m<  ridian  plane  is  revolved  to  its  true 
position,  X  will  be  horizon ta  ;y  projected  in  mn  at  s;  sc  will  be 
the  horizontal  trace  of  the  meridian  plane,  and  u  will  be  the 
horizontal,  and  u'  the  vertical  projection  of  the  point  of  contac 
on  the  given  surface.  A  plane  through  this  point  and  MN  will 
be  the  tangent  plane. 


DESCRIPTIVE   GEOMETRY.  85 

153.  In  general,  through  a  given  right  line  a  limited  number 
of  planes  only  can  be  passed  tangent  to  a  double  curved  surface. 
For  let  the  surface  be  intersected  by  a  system  of  planes  parallel 
to  the  given  line,  and  tangents  be  drawn  to  the  sections  also 
parallel  to  the  line.  These  will  form  a  cylinder  tangent  to  the 
surface.  Any  plane  through  the  right  line  tangent  to  this  cylin- 
der will  be  tangent  to  the  surface ;  and  the  number  of  tangent 
planes  will  be  determined  by  the  number  of  tangents  which  can 
be  drawn  from  a  point  of  the  given  line  to  a  section  made  by  a 
plane  through  this  point. 


POINTS    IN    WHICH    SURFACES    ARE    PIERCED    BY    LINES.  , 

154.  The  points  in  which  a  right  line  pierces  a  surface  are 
easily  found  by  passing  through  the  line  any  plane  intersecting 
the  surface.  It  will  cut  from  the  surface  a  line,  which  will  iuter- 
sect  the  given  line  in  the  required  points.  This  auxiliary  plane 
should  be  so  chosen  as  to  intersect  the  surface  in  the  simplest  line 
possible. 

If  the  given  surface  be  a  cylinder,  a  plane  through  the  right 
line  parallel  to  the  rectilinear  elements  should  be  used.  It  will 
intersect  the  cylinder  in  one  or  more  rectilinear  elements,  which 
will  intersect  the  given  line  in  the  required  points. 

If  the  given  surface  be  a  cone,  the  auxiliary  plane  should  pass 
through  the  vertex. 

If  a  sphere,  the  auxiliary  plane  should  pass  through  the  centre, 
as  the  line  of  intersection  will  be  a  circumference  with  a  radius 
equal  to  that  of  the  sphere. 


155.   If  the  given  line  be  a  curve  of  single  curvature,  its  plane 
will  intersect  the  surface,  if  at  all,  in  a  line  which  will  contain  the 
required  points.     If  the  plane  does  not  intersect  the  surface,  or  if  . 
the  line  of  intersection  does  not  intersect,  or  is  not  tangent  to  the 


86  DESCRIPTIVE   GEOMETRY. 

given  curve,  there  will,  of  course,  be  no  points  common  to  the 
curve  and  surface. 

If  the  given  line  be  a  curve  of  double  curvature,  a  cylinder  or 
cone  may  be  passed  through  it  intersecting  the  given  surface.  If 
the  line  of  intersection  intersects  the  given  line,  the  points  thus 
determined  will  be  the  required  points. 

These  problems  will  be  easily  solved  when  the  more  general 
problem  of  finding  the  intersection  of  surfaces  has  been  discussed. 


INTERSECTION     OF     SURFACES     BY     PLANES.        DEVELOPMENT     OF 
SINGLE     CURVED     SURFACES. 

156.  The  solution  of  the  problem  of  the  intersection  of  sur- 
faces consists  in  finding  two  lines,  one  on  each  surface,  which  in- 
tersect. The  points  of  intersection  will  be  points  in  both  surfaces, 
and  therefore  points  of  their  line  of  intersection. 


157.  To  find  the  intersection  of  a  plane  with  a  surface,  we 
intersect  the  plane  and  surface  by  a  system  of  auxiliary  planes. 
Each  plane  will  cut  from  the  given  plane  a  right  line,  and  from 
the  surface  a  line,  the  intersection  of  which  will  be  points  of  the 
required  line.  The  system  of  auxiliary  planes  should  be  so  chosen 
as  to  cut  from  the  surface  the  simplest  lines;  a  rectilinear  ele- 
ment, if  possible,  or  the  circumference  of  a  circle,  &c. 

The  curve  of  intersection  may  be  drawn  with  greater  accuracy 
by  determining  at  each  of  the  points  thus  found  a  tangent  to 
the  curve,  and  then  drawing  the  projections  of  the  curve  tangent 
to  the  projections  of  these  tangents,  Art.  (65). 

This  tangent  must  lie  in  the  intersecting  plane,  that  is,  in  the 
plane  of  the  curve,  Art.  (64).  It  must  also  lie  in  the  tangent 
plane  to  the  surface  at  the  given  point,  Art.  (108).  Hence,  if 
we  pass  a  plane  tangent  to  the  given  surface  at  the  given  point,  and 


DESCRIPTIVE   GEOMETRY.  87 

determine  its  intersection  with  the  intersecting  plane  ;  this  will  be 
the  required  tangenU 


158»  Since  the  tangent  plane  to  a  single  curved  surface  in 
general  contains  two  consecutive  rectilinear  elements,  Art.  (Ill), 
it  will  contain  the  elementary  portion  of  the  surface  generated 
by  the  generatrix  in  moving  from  the  first  element  to  the  second. 
Now  if  the  surface  be  rolled  over  until  the  consecutive  element 
following  the  second  comes  into  the  tangent  plane,  the  portion  oi 
the  plane  limited  by  the  first  and  third  elements  will  equal  the 
portion  of  the  surface  limited  by  the  same  elements.  If  we  con- 
tinue to  roll  the  surface  on  the  tangent  plane  until  any  following 
element  comes  into  it,  the  portion  of  the  plane  included  between 
this  element  and  the  first  will  be  equal  to  the  portion  of  the  sur- 
face limited  by  the  same  elements.  Therefore  if  a  single  curved 
surface  be  rolled  over  on  any  one  of  its  tangent  planes  until  each 
of  its  rectilinear  elements  has  come  into  this  plane,  the  portion  of 
the  plane  thus  touched  by  the  surface,  and  limited  by  the  extreme 
elements,  will  be  a  plane  surface  equal  to  the  given  surface,  and  is 
the  development  of  the  surface. 

As  a  tangent  plane  to  a  warped  surface  cannot  contain  two 
consecutive  rectilinear  elements,  Art.  (113),  the  elementary  sur- 
face limited  by  these  two  elements  cannot  be  brought  into  a  plane 
without  breaking  the  continuity  of  the  surface.  A  warped  sur- 
face, therefore,  cannot  be  developed* 

Neither  can  a  double  curved  surface  be  developed ;  as  any  ele- 
mentary portion  of  the  surface  will  be  limited  by  two  curves,  and 
cannot  be  brought  into  a  plane  without  breaking  the  continuity  of 
the  surface. 


159,  In  order  to  determine  the  position  of  the  different  recti- 
linear elements  of  a  single  curved  surface,  as  they  come  into  the 
tangent  plane,  ov  plane  of  development,  it  will  always  be  necessary 


88  DESCRIPTIVE   G-EOMETKY. 

to  find  some  curve  upon  the  surface  which  will  develop  into  a 
right  line,  or  circle,  or  some  simple  known  curve,  upon  which  the 
rectified  distances  between  these  elements  can  be  laid  off. 


160.  Problem  37.  To  find  the  intersection  of  a  right  cylinder 
with  o  circular  base  by  a  plane. 

Let  mlo,  Fig.  (64),  be  the  base  of  the  cylinder,  and  c  the  hori- 
zontal, and  c'd'  the  vertical  projection  of  the  axis;  then  mlo  will 
be  the  horizontal,  and  s'm"o"u'  the  vertical  projection  of  the 
cylinder,  Art.  (75).     Let  tTt'  be  the  intersecting  plane. 

Analysis.  Intersect  the  cylinder  and  plane  by  a  system  of 
auxiliary  planes  parallel  to  the  axis,  and  also  to  the  horizontal 
trace  of  the  given  plane.  These  will  cut  from  the  cylinder  recti- 
linear elements,  and  from  the  plane  right  lines  parallel  to  the 
horizontal  trace.  The  intersection  of  these  lines  will  be  points 
of  the  required  curve,  Art.  (157). 

This  curve,  as  also  the  intersection  of  any  cylinder  or  cone 
with  a  circular  or  an  elliptical  base,  by  a  plane  cutting  all  the 
rectilinear  elements,  is  an  ellipse.  Analyt.  Geo.,  Arts.  (82  and 
200). 

Construction.  Draw  xy  parallel  to  t%  it  will  be  the  horizon- 
tal trace  of  one  of  the  auxiliary  planes ;  yy'  is  its  vertical  trace. 
It  intersects  the  cylinder  in  two  elements,  one  of  which  pierces 
H  at  x,  and  the  other  at  z,  and  x"x'  and  z"z'  are  their  vertical 
projections.  It  intersects  the  plane  in  the  right  line  XY,  Art. 
(38),  and  X  and  Z,  the  intersections  of  this  line  with  the  two 
elements,  are  points  of  the  curve.  In  the  same  way  any  number 
of  points  may  be  found. 

If  a  point  of  the  curve  on  any  particular  element  be  required, 
we  have  simply  to  pass  an  auxiliary  plane  through  this  element. 
Thus,  to  construct  the  point  on  (o,  o"u'),  draw  -the  trace  ow,  and 
construct  as  above  the  point  0. 

Since  this  point  lies  on  the  curve,  and  also  on  the  extreme 
element,  and  since  no  point  of  the  curve  can  be  vertically  pro- 


DESCRIPTIVE    GEOMETRY.  89 

jected  outside  of  o"u',  the  vertical  projection  of  the  curve  must 
be  tangent  to  o"u'  at  o'.  Or,  the  reason  may  be  given  thus,  in 
many  like  cases:  If  a  tangent  be  drawn  to  the  curve  at  O,  it  will 
lie  in  the  tangent  plane  to  the  surface  at  O,  Art.  (108) ;  and  since 
this  tangent  plane  is  perpendicular  to  the  vertical  plane,  the  tan- 
gent will  be  vertically  projected  into  its  trace  o"u\  which  must 
therefore  be  tangent  to  m'o'V  at  o',  Art.  (65).  Also,  m"s'  must 
be  tangent  to  m'o'V  at  m'. 

If  a  plane  be  passed  through  the  axis  perpendicular  to  the  in- 
tersecting plane,  it  will  evidently  cut  from  the  plane  a  right  line, 
which  will  bisect  all  the  chords  of  the  curve  perpendicular  to  it, 
and  this  line  will  be  the  transverse  axis  of  the  ellipse.  Ic  is  the 
horizontal  trace  of  such  a  plane.  It  cuts  the  given  plane  in  KG, 
Art.  (38),  and  the  cylinder  in  two  elements  horizontally  projected 
at  k  and  /,  and  KL  is  the  transverse  axis,  and  K  and  L  the  ver- 
tices of  the  ellipse.  K  is  the  lowest,  and  L  the  highest  point  of 
the  curve. 

Since  the  curve  lies  on  the  surface  of  the  cylinder,  its  hori- 
zontal projection  will  be  in  the  base  mlo. 

To  draw  a  tangent  to  ike  curve  at  any  point,  as  X ;  draw  xr 
tangent  to  xol  at  x ;  it  is  the  horizontal  trace  of  a  tangent  plane  to 
the  cylinder  at  X,  Art.  (123).  This  plane  intersects  tTV  in  a 
right  line,  which  pierces  H  at  r;  and  since  X  is  also  a  point  of 
the  intersection,  RX  will  be  the  required  tangent,  Art.  (157).  If 
a  tangent  be  drawn  at  each  of  the  points  determined  as  above, 
the  projections  of  the  curve  can  be  drawn,  with  great  accuracy, 
tangent  to  the, projections  of  these  tangents,  at  the  projections  of 
the  points  of  tangency. 

The  part  of  the  curve  MXO,  between  the  points  M  and  O,  lies 
in  front  of  the  extreme  elements  (m,  m"s')  and  (o,  o"u'),  and  y 
seen,  and  therefore  m'x'o'  is  drawn  full,  and  m'l'o'  broken. 

To  represent  the  curve  in  its  true  dimensions  let  its  plane  be  re- 
volved about  tT  until  it  coincides  with  H.  The  revolved  position 
of  each  point,  as  X?  at  x'",  will  be  determined  as  in  Art  (17)~ 
k'"  and  V"  will  be  the  revolved  position  of  the  vertices,  rx'"  will 


90  DESCRIPTIVE    GEOMETRY. 

be  the  revolved  position  of  the  tangent,  and  V"x'"kHf  the  ellipse 
in  its  true  size. 


161.  Problem  38.  To  develop  a  right  cylinder  with  a  circular 
base,  and  trace  upon  the  development  the  curve  of  intersection  of  the 
cylinder  by  an  oblique  plane. 

Let  the  cylinder  and  curve  be  given  as  in  the  preceding  prob- 
lem, and  let  the  plane  of  development  be  the  tangent  plane  at  L. 

Analysis.  Since  the  plane  of  the  base  is  perpendicular  to  the 
element  of  contact  of  the  tangent  plane,  it  is  evident  that  as  the 
cylinder  is  rolled  out  on  this  plane,  this  base  will  develop  into  a 
right  line,  on  which  we  can  lay  off  the  rectified  distances  between 
the  several  elements,  and  then  draw  them  each  parallel  to  the 
element  of  contact. 

Points  of  the  development  of  the  curve  are  found  by  laying 
off  on  the  development  of  each  element,  from  the  point  where  it- 
meets  the  rectified  base,  a  distance  equal  to  the  distance  of  the 
point  from  the  base. 

Construction.  The  plane  of  development  being  coincident 
with  the  plane  of  the  paper,  let  ZL,  Fig.  65,  be  the  element  of 
contact.  Draw  //  perpendicular  to  /L,  and  lay  off  //  equal  to  the 
rectified  circumference  lw--k -  -I,  Fig.  64.  It  will  be  the  develop- 
ment of  the  base.  Lay  off  Iw  equal  to  the  arc  lwy  and  draw  wW 
parallel  to  lh.  It  is  the  development  of  the  element  which 
pierces  II  at  w.  Likewise  for  each  of  the  elements  lay  off  wz, 
zm,  &c,  equal  respectively  to  the  rectified  arcs  wzy  zm,  &c,  in 
Fig.  64,  and  draw  zZ,  mM,  &c.  The  portion  of  the  plane  included 
between  lh  and  lh  will  be  the  development  of  the  cylinder. 

On  lh  lay  off  lh  =  l"l',h  will  be  one  point  of  the  developed 
curve ;  on  wW  lay  off  wW  =  w"w\  W  will  be  a  second  point ; 
and  thus  each  point  may  be  determined,  and  L--K--L  will  be 
the  developed  curve.  The  line  rx,  the  sub-tangent,  will  take  the 
position  rx  on  the  developed  base,  and  Xr  will  be  the  tangent  at 
X  in  the  plane  of  development.      This  must  be  tangent  at  X, 


DESCRIF1TVE   GEOMETRY.  91 

since  the  tangent  after  development  must  contain  the  same  two 
consecutive  points  which  it  contains  in  space,  and  therefore  be 
tangent  to  the  development  of  the  curve,  Art.  (64). 


162.  Problem  39.  To  find  the  intersection  of  an  oblique  cylin- 
ier  by  a  plane. 

Let  the  cylinder  be  given  as  in  Fig.  66,  and  let  tTt',  perpen- 
dicular to  the  rectilinear  elements,  be  the  intersecting  plane. 

Analysis.  Intersect  the  cylinder  and  plane  by  a  system  of 
auxiliary  planes  parallel  to  the  rectilinear  elements,  and  perpen- 
dicular to  the  horizontal  plane.  These  planes  will  each  intersect 
the  cylinder  in  two  rectilinear  elements,  and  the  plane  in  a  right 
line,  the  intersection  of  which  will  Be  points  of  the  curve 

Construction.  Let  eg,  parallel  to  ft,  be  the  horizontal  trace  of 
an  auxiliary  plane ;  qq'  will  be  its  vertical  trace.  It  intersects  the 
cylinder  in  two  elements,  one  of  which  pierces  H  at  r,  the  other 
at  s,  and  these  are  vertically  projected  in  r'g'  and  s'h',  and  hori- 
zontally in  eq.  It  intersects  tTt'  in  a  right  line,  which  pierces  H 
at  e,  and  is  vertically  projected  in  e'q',  and  horizontally  in  eq. 
These  lines  intersect  in  Z  and  Y,  Art.  (23),  points  of  the  curve. 
In  the  same  way  any  number  of  points  may  be  determined. 

The  auxiliary  planes,  being  parallel,  must  intersect  tTt'  in 
parallel  lines,  the  vertical  projections  of  which  will  be  parallel 
to  e'q'. 

By  the  plane,  whose  horizontal  trace  is  mn,  the  points  U  and 
X  are  determined.  The  vertical  projection  of  the  curve  must  be 
tangent  to  m'n'  at  u'.  The  points  in  which  the  horizontal  pro- 
jection is  tangent  to  li  and  kf,  are  determined  by  using  these 
lines  as  the  traces  of  auxiliary  planes. 

To  draw  a  tangent  to  the  curve  at  any  point,  as  X,  pass  a  plan 
tangent  to  the  cylinder  at  X ;  av  is  its  horizontal  trace,  and  vx  th 
horizontal,  and  v'x'  the  vertical  projection  of  the  tangent. 

A  sufficient  number  of  points  and  tangents  being  thus  deter 
mined,  the  projections  of  the  curve  can  be  drawn  with  accuracy 


92  DESCRIPTIVE   GEOMETRY. 

The  part  cyd  is  full,  being  the  horizontal  projection  of  that  part 
of  the  curve  which  lies  above  the  extreme  elements  LI  and  KF. 
For  a  similar  reason,  u'x'w'  is  full. 

To  show  the  curve  in  its  true  dimensions,  revolve  the  plane 
about  tT  until  it  coincides  with  H.  The  revolved  position  of  each 
point  may  be  found  as  in  Art.  (17),  and  c"y"d"z"  will  be  the 
curve  in  its  true  size. 

The  section  thus  made  is  a  right  section. 


163.  If  it  be  required  to  develop  the  cylinder  on  a  tangent  plane 
along  any  element,  as  KF,  we  first  make  a  right  section  as  above. 
We  know  this  will  develop  into  a  right  line  perpendicular  to  KF. 
On  this  we  lay  off  the  rectified  arcs  of  Ihe  section  included  be- 
tween the  several  elements,  and  then  draw  these  elements  paral- 
lel to  KF. 

The  developed  base,  or  any  curve  on  the  surface,  may  be  traced 
on  the  plane  of  development  by  laying  off  on  each  element,  from 
the  developed  position  of  the  point  where  it  intersects  the  right 
section,  the  distance  from  this  point  to  the  point  where  the  ele- 
ment intersects  the  base  or  curve.  A  line  through  the  extremities 
of  these  distances  will  be  the  required  development. 


164.  Problem  40.  To  find  the  intersection  of  a  right  cone  icith 
a  circular  base  by  a  plane. 

Let  the  cone  be  given  as  in  Fig.  67,  and  let  tTt'  be  the  given 
plane,  the  vertical  plane  being  assumed  perpendicular  to  it. 

Analysis.  Intersect  the  cone  by  a  system  of  planes  through 
the  vertex  and  perpendicular  to  the  vertical  plane.  The  elements 
cut  from  the  cone  by  each  plane,  Art.  (157),  will  intersect  the 
right  line  cut  from  the  given  plane  in  points  of  the  required 
curve. 

Constructor.  Let  lie  be  the  horizontal  trace  of  an  auxiliary 
plane,  ks'  will  be  its  vertical  trace.     It  intersects  the  cone  in  two 


DESCRIPTIVE   GEOMETRY.  03 

elements,  one  of  which  pierces  H  in  /,  and  the  other  in  i,  hori- 
zontally projected  in  h  and  is  respectively.  It  intersects  the 
given  plane  in  a  right  line  perpendicular  to  V,  vertically  projected 
at  x,  and  horizontally  in  xv ;  hence  x  and  v  are  the  horizontal 
projections  of  two  points  of  the  curve,  both  vertically  projected  at 
x '.  In  the  same  way  any  number  of  points  can  be  found,  as  Y 
(wy'),  &c. 

The  plane  whose  horizontal  trace  is  mo,  perpendicular  to  tTt', 
intersects  it  in  a  right  line  vertically  projected  in  Tt',  which  evi- 
dently bisects  all  chords  of  the  curve  perpendicular  to  it,  and  is 
therefore  an  axis  of  the  curve.  This  plane  cuts  from  the  cone 
the  elements  SM  and. SO,  which  are  intersected  by  the  axis  in  the 
points  Z  and  U,  which  are  the  vertices. 

To  draw  a  tangent  to  the  curve  at  any  point,  as  X,  pass  a  plane 
tangent  to  the  cone  at  X,  Ir  is  its  horizontal  trace.  It  intersects 
tTt'  in  (rx,  Tx'),  which  is  therefore  the  required  tangent,  Art. 
(157).  The  horizontal  projection,  uxzv,  can  now  be  drawn,  u'z' 
is  its  vertical  projection. 

To  represent  the  curve  in  its  true  dimensions,  we  may  revolve  it 
about  ^T  until  it  coincides  with  H,  or  about  T^'  until  it  coincides 
with  V,  and  determine  it  as  in  Art.  (17).  Otherwise,  thus  :  Re- 
volve it  about  UZ  until  its  plane  becomes  parallel  to  V,  it  will  then 
be  vertically  projected  in  its  true  dimensions,  Art.  (62).  The 
points  U  and  Z  being  in  the  axis,  will  be  projected  at  u'  and  z' 
respectively.  X  will  be  vertically  projected  at  x",  x'x"  being 
equal  to  px.  Y  at  y" ,  y'y"  being  equal  to  qy.  (wy')  at  w",  &c, 
and  u'x"z'x'"  will  be  the  curve  in  its  true  size. 


165.  If  a  right  cone,  with  a  circular  base,  be  intersected  by  a 
plane,  as  in  Fig.  67,  making  a  less  angle  with  the  plane  of  the 
base  than  the  elements  do,  the  curve  of  intersection  is  an  ellipse 
Analyt.  Geo ,  Art.  (82.)  If  it  make  the  same  angle,  or  is  paralle 
to  one  of  the  elements,  the  curve  is  a  parabola.     If  if  *«ake    . 


9Jr  DESCRIPTIVE   GEOMETRY. 

greater  angle,  the  curve  is  an  hyperbola.      Hence   these  three 
curves  are  known  by  the  general  name,  conic  i 


166.  Problem  41.  To  develop  a  right  cone  with  a  circular 
base. 

Let  the  cone  and  its  intersection  by  an  oblique  plane  be  given 
as  in  the  preceding  problem,  Fig.  6*7,  and  let  the  plane  of  develop- 
ment be  the  tangent  plane  along  the  element  MS;  the  half  of  the 
cone  in  front  being  rolled  to  the  left,  and  the  other  half  to  the 
right. 

Analysis.  Since  the  base  of  the  cone  is  everywhere  equally 
distant  from  the  vertex,  as  the  cone  is  rolled  out,  each  point  of 
this  base  will  be  in  the  circumference  of  a  circle  described  with 
the  vertex  as  a  centre,  and  a  radius  equal  to  the  distance  from  the 
vertex  to  any  point  of  the  base.  By  laying  off  on  this  circum- 
ference the  rectified  arc  of  the  base,  contained  between  any 
two  elements,  and  drawing  right  lines  from  the  extremities  to  the 
vertex,  we  have,  in  the  plane  of  development,  the  position  of 
these  elements.  Laying  off  on  the  proper  elements  the  distances 
from  the  vertex  to  the  different  points  of  the  curve  of  intersec- 
tion, and  tracing  a  curve  through  the  extremities,  we  have  the 
development  of  the  curve  of  intersection. 

Construction.  With  SM  =  s'm\  Figs.  67  and  68,  describe  the 
arc  OMO.  It  is  the  development  of  the  base.  Lay  off  MG  = 
mg,  and  draw  SG ;  it  is  the  developed  position  of  the  element  SG. 
In  the  same  way  lay  off  GL  =  gl,  LO  =  lo,  and  draw  SL  and 
SO.  OSM  is  the  development  of  one-half  the  cone.  In  like 
manner  the  other  half  may  be  developed. 

On  SM  lay  off  SZ  —  srz'.  Z  will  be  the  position  of  the  point  Z 
in  the  plane  of  development.  To  obtain  the  true  distance  from 
S  to  Y,  revolve  SY  about  the  axis  of  the  cone  until  it  becomes 
parallel  to  V,  as  in  Art.  (28) ;  s'e'  will  be  its  true  length.  On  SG 
lay  off  SY  =  s'e' ;  on  SL,  SX  =  s'd' ;  on  SM,  SU  =  s'a'.  Z,  Y,  X, 
and  U  will  be  the  position    of  these   points   on    the    plane    oi 


DESCRIPTIVE   GEOMETRY.  05 

development,  and   TJX---U   will   be    the   development  of   the 
curve  of  intersection. 

Through  L  draw  LR  perpendicular  to  LS,  and  make  it  equal  to 
//•.     RX  will  be  the  developed  tangent. 


167.  Problem  42.  To  find  the  intersection  of  any  cone  by  a 
plane. 

Let  the  cone  and  plane  fit'  be  given  as  in  Fig.  69. 

Analysis.  Intersect  the  cone  by  a  system  of  planes  through 
the  vertex  and  perpendicular  to  the  horizontal  plane.  Each  of 
these  planes  will  intersect  the  cone  in  one  or  more  rectilinear 
elements,  and  the  given  plane  in  a  right  line,  the  intersection  of 
which  will  be  points  of  the  curve.  Since  these  auxiliary  planes 
are  perpendicular  to  the  horizontal  plane,  they  will  intersect  in  a 
right  line  through  the  vertex,  perpendicular  to  the  horizontal 
plane,  and  the  point  in  which  this  line  pierces  the  cutting  plane 
will  be  a  point  common  to  ail  the  right  lines  cut  from  this  plane. 

Construction.  Find  the  point  in  which  the  perpendicular  to 
H,  through  S,  pierces  tTt'r  as  in  Art.  (42).  v'  is  its  vertical  pro- 
jection. The  vertical  projections  of  the  lines  cut  from  tTt'  all 
pass  through  this  point. 

Let  sp  be  the  horizontal  trace  of  an  auxiliary  plane.  It  inter- 
sects the  cone  in  the  elements  SE  and  SD,  and  the  cutting  plane 
in  the  right  line  (ps,  p'v').  This  line  intersects  the  elements  in 
R  and  Y,  which  are  points  of  the  required  curve.  In  the  same 
way  any  number  of  points  may  be  found. 

To  find  the  point  of  the  curve  on  any  particular  element,  as 
SM,  we  pass  an  auxiliary  plane  through  this  element,  sm  is  its 
horizontal  trace,  and  Z  and  X  the  two  points  on  this  element. 
The  vertical  projection  of  the  curve  is  tangent  to  s'm'  at  z',  and 
to  sV  at  u' .  The  points  q  and  w,  in  which  the  horizontal  pro- 
jection is  tangent  to  si  and  sn,  are  found  by  using  as  auxiliary 
planes  the  two  planes  whose  traces  are  si  and  sn. 

To  draw  a  tangent  to  the  curve  at  any  point,  as  X,  pass  a  plane 


96  PESCKIPTIVK  GEOMETRY. 

tangent  to  the  cone  at  X.  ic  is  its  horizontal  trace,  Art.  (129). 
It  intersects  fit'  in  cX,  which  is  therefore  the  required  tangent, 
Art.  (157). 

The  part  of  the  curve  which  lies  above  the  two  extreme  ele- 
ments SL  and  SN,  is  seen,  and  therefore  its  projection,  wyzq,  is 
full.  For  a  similar  reason  the  projection,  z'q'x'u',  of  that  part  of 
the  curve  which  lies  in  front  of  the  two  extreme  elements,  SM 
and  SO,  is  full. 

To  show  the  curve  in  its  true  dimensions,  revolve  the  plane 
about  tT  until  it  coincides  with  H,  and  determine  each  point,  as 
Q  at  q",  as  in  Art.  (17).  Or  the  position  of  (svr)  may  be  found 
at  v".  Then  if  the  points  c,  «,  6,  p,  &c.,  be  each  joined  with  this 
point  by  right  lines,  we  have  the  revolved  positions  of  the  lines 
cut  from  the  given  plane  by  the  auxiliary  planes,  and  the  points 
y",  z" ,  r' ',  x",  in  which  these  lines  are  intersected  by  the  perpen- 
diculars to  the  axis,  yy" ',  zz",  &c,  are  points  of  the  revolved  po- 
sition of  the  curve.     x"c  is  the  revolved  position  of  the  tangent. 


168.  The  intersection  of  the  single  curved  surface,  with  a 
helical  directrix,  by  a  plane,  may  be  found  by  intersecting  by  a 
system  of  auxiliary  planes  tangent  to  the  projecting  cylinder  of 
the  helical  directrix.  These  intersect  the  surface  in  rectilinear 
elements,  and  the  plane  in  right  lines,  the  intersection  of  which 
will  be  points  of  the  required  curve. 


169.  Problem  43.  To  find  the  intersection  of  any  surface  of 
revolution  by  a  plane. 

Let  the  surface  be  a  hyperboloid  of  revolution  of  one  nappe, 
given  as  in  Fig.  70,  and  let  tit'  be  the  cutting  plane. 

Analysis.  If  a  meridian  plane  be  passed  perpendicular  to.  the 
cutting  plane,  it  will  intersect  it  in  a  right  line,  which  will  divide 
the  curve  symmetrically,  and  be  an  axis.  If  the  points  in  which 
t.his  line  pierces  the  surface  be  found,  these  will  be  the  vertices  of 


DESCRIPTIVE    GEOMETRY.  97 

the  curve.  Now  intersect  by  a  system  of  planes  perpendicular  to 
the  axis  of  the  surface ;  each  plane  will  cut  from  the  surface  a  cir- 
cumference and  from  the  given  plane  a  right  line,  the  intersection 
of  which  will  be  points  of  the  required  curve. 

Construction.  Draw  en  perpendicular  to  trY1  it  will  be  the  hori- 
zontal trace  of  the  auxiliary  meridian  plane.  This  plane  inter- 
sects tTt'  in  the  right  line  NC,  Art.  (38),  and  the  surface  in  a 
meridian  curve  which  is  intersected  by  NC  in  the  two  vertices  of 
the  required  curve ;  or  in  its  highest  and  lowest  points. 

To  find  these  points,  revolve  the  meridian  plane  about  the  axis 
until  it  becomes  parallel  to  V.  The  meridian  curve  will  be  verti- 
cally projected  into  the  hyperbola,  which  limits  the  ver.ical  pro- 
jection of  the  surface,  Art.  (02),  and  the  line  NC  into  d'c'.  The 
poims  s'  and  r'  will  be  the  vertical  projections  of  the  revolved  po- 
sitions of  the  vertices.  After  the  counter  revolution,  these  points  are 
horizontally  projected  at  k  and  /,  and  vertically  at  k'  and  V . 

Let  u'z  be  the  vertical  trace  of  an  auxiliary  plane  perpendicular 
to  the  axis.  It  cuts  the  surface  in  a  circle  horizontally  projected 
in  wuz,  aud  the  plane  in  a  right  line,  which  piercing  V  at  e'  is 
horizontally  projected  in  ez ;  hence  U  and  Z  are  points  of  the 
curve.  Id  the  same  way  any  number  of  points  may  be  deter- 
mined. 

The  points  o'  and  v\  in  which  the  vertical  projection  of  the 
curve  is  tangent  to  the  hyperbola  which  limits  the  projection  of 
the  surface,  are  the  vertical  projections  of  the  points  in  which  the 
line,  cut  out  of  the  given  plane  by  the  meridian  plane  parallel  to 
the  vertical  plane,  intersects  the  meridian  curve  cut  out  by  the 
same  plane. 

The  points  upon  any  particular  circle  may  be  determined  by 
using  the  plane  of  this  circle  as  an  auxiliary  plane.  If  the  curve 
crosses  the  circle  of  the  gorge,  the  points  in  which  it  crosses  are 
determined  by  using  the  plane  of  this  circle,  and,  in  this  case,  the 
horizontal  projection  of  the  curve  must  be  tangent  to  the  horizon- 
tal projection  of  the  circle  of  the  gorge,  at  the  points  x  and  y. 

The  method  given  above  for  determining  the  vertices  of  the 


98  DESCRIPTIVE    GEOMETRY. 

curve  is  applicable  to  any  surface  of  revolution.  In  this  particular 
surface  it  may  be  modified  thus  :  Let  the  line  NL  revolve  about 
the  axis;  it  will  generate  a  cone  of  revolution  whose  base  is  dni^ 
and  vertex  Ci  This  cone  intersects  the  hyperboloid  in  two  cir- 
cumferences, Art.  (97),  and  the  points  in  which  NL  intersects 
these  circumferences  will  be  the  points  required. 

To  construct  them  ;  through  any  rectilinear  element  of  the 
hyperbol<5id,  as  MQ,  and  the  vertex  of  the  cone^  pass  a  plane ;  qf 
will  be  its  horizontal  trace,  Art.  (33).  It  intersects  the  cone  in 
two  elements  horizontally  projected  in  ci  and  cg^  and  the  points  a 
and  6  are  points  in  the  horizontal  projections  of  the  circles  in 
which  the  hyperboloid  and  cone  intersect,  and  K  and  L  are  the 
required  vertices. 

To  draw  a  tangent  to  the  curve  at  Z,  pass  a  plane  tangent  to  the 
surface  at  Z,  as  in  Art.  (138);  its  intersection  with  tTt'  will  be  the 
tangent,  Art.  (157). 

The  curve  may  be  represented  in  its  true  dimensions  as  in  Art. 
(162). 

Let  the  intersection  of  an  oblique  plane  with  a  sphere,  an  ellip- 
soid of  revolution  and  paraboloid  of  revolution,  be  constructed  in 
accordance  with  the  principles  of  the  preceding  problem. 


170.  To  find  the  intersection  of  any  warped  surface  with  a 
plane  directer,  by  an  oblique  plane,  intersect  by  planes  parallel  to 
the  plaue  directer.  Each  will  cut  from  the  surface  one  or  more 
rectilinear  elements,  and  from  the  plane  a  right  line,  the  intersec- 
tion of  which  will  be  points  of  the  required  curve. 


171.  To  find  the  intersection  qf  a  helicoid  by  a  plane,  the  sur- 
face being  given  as  in  Art.  (92) ;  intersect  by  a  system  of  auxiliary 
planes  through  the  axis.  These  will  cut  from  the  surface  recti- 
linear elements,  and  from  the  plane  right  lines,  the  intersection  of 
which  will  be  points  of  the  required  curve. 


DESCRIPTIVE   GEOMETRY.  99 

Let  the  construction  be  made,  and  the  curve  and  its  tangent 
represented  in  true  dimensions. 


INTERSECTION  OF  CURVED  SURFACES. 

172.  To  find  the  intersection  of  any  two  curved  surfaces,  we  in- 
tersect them  by  a  system  of  auxiliary  surfaces^  Each  auxiliary 
surface  will  cut  from  the  given  surfaces  lines  the  intersection  of 
which  will  be  points  of  the  required  line.     Art.  (156). 

The  system  of  auxiliary  surfaces  should  be  so  chosen  a?  to  cut 
from  the  given  surfaces  the  simplest  lines,  rectilinear  elements  if 
possible,  or  the  circumferences  of  circles,  &c. 

To  draw  a  tangent  to  the  curve  of  intersection  at  any  point ;  pass 
a  plane  tangent  to  each  surface  at  this  point.  The  intersection  of 
these  two  planes  will  be  the  required  tangent,  since  it  must  lie  in 
each  of  the  tangent  planes.     Art.  (108). 

In  constructing  this  curve  of  intersection,  great  care  should  be 
taken  to  determine  those  points  in  which  its  projections  are  tan- 
gent to  the  limiting  lines  of  the  projections  of  the  surfaces ;  and 
also  those  points  in  which  the  curve  itself  is  tangent  to  other  lines 
of  either  surface,  as  these  points  aid  much  in  drawing  the  curve 
with  accuracy. 


173.  Problem  44.  To  find  the  intersection  of  a  cylinder  and 
cone. 

Let  the  surfaces  be  given  as  in  Fig.  71 ;  the  base  of  the  cylin- 
der bai  being  in  the  horizontal  plane,  and  its  rectilinear  elements 
parallel  to  the  vertical  plane ;  the  base  of  the  cone  m'l'o',  in  the 
vertical  plane,  and  S  its  vertex. 

Analysis.  Intersect  the  two  surfaces  by  a  system  of  auxiliary 
planes  passing  through  the  vertex  of  the  cone  and  parallel  to  the 
rectilinear  elements  of  the  cylinder.  Each  plane  will  intersect  each 
of  the  surfaces  in  two  rectilinear  elements,  the  intersection  of  which 


100  DESCRIPTIVE    GKOMKTKY. 

wiM  be  points  of  the  required  curve.  These  planes  will  intersect 
in  a  right,  line  passing  through  the  vertex  of  the  cone  and  parallel 
to  the  rectilinear  elements  of  the  cylinder,  and  the  point,  in  which 
this  line  pierces  the  horizontal  plane,  will  be  a  point  common  to 
the  horizontal  traces  of  all  the  auxiliary  planes. 

Construction.  Through  S  draw  ST  parallel  to  BE.  It  pierces 
FI  in  t.  Through  t  draw  any  right  line,  as  th]  it  may  be  taken  as 
the  horizontal  trace  of  an  auxiliary  plane,  the  vertical  traca  of  which 
is  hg'  parallel  to  b'e'.  This  plane  intersects  the  cylinder  in  two 
elements  which  pierce  II  at  g  and  t,  and  are  horizontally  projected 
in  gg"  and  iin.  The  same  plane  intersects  the  cone  in  two  ele- 
ments which  pierce  V  in  m'  and  rt,  and  are  horizontally  pro- 
jected in  ms  and  its.  These  elements  intersect  in  the  points  X,  Y 
Z  and  U,  which  are  points  of  the  required  curve.  In  the  same 
way  any  number  of  points  may  be  determined. 

The  horizontal  projection  of  the  required  curve  is  tangent  to 
ms  at  the  points  x  and  // ;  since  ms  is  the  horizontal  projection  of 
one  of  the  extreme  elements  of  the  cone.  The  points  of  tangency 
on  si  may  be  determined  by  using  an  auxiliary  plane,  which  shall 
contain  the  element  SL. 

The  points  of  tangency  on  dw  are  obtained  in  the  same  way,  by 
using  the  plane  whose  horizontal  trace  is  td. 

The  points  in  which  the  vertical  projection  of  the  curve  is  tan- 
„gent  to  the  vertical  projections  of  the  extreme  elements  of  both 
cone  and  cylinder  will  be  determined,  by  using  as  auxiliary  planes 
those  which  contain  these  elements. 

A  tangent  to  the  curve  at  any  of  the  points,  thus  determined, 
may  be  constructed  by  finding  the  intersection  of  two  planes,  one 
tangent  to  the  cylinder  and  ihe  other  to  the  cone,  at  this  point, 
Art.  (172). 

The  plane,  of  which  tc  is  the  horizontal  trace,  is  tangent  to  the 
cylinder  along  the  element  CQ,  and  intersects  the  cone  in  the  two 
elements  SP  and  SQ.     If  at  P  a  plane  be  passed  tangent  to  th 
cone,  it  will  be  tangent  along  SP,  which  is  also  its  intersection 
with  the  tangent  plane  to  the  cylinder.     SP  is  then  tangent  to 


DESCRIPTIVE   GEOMETRY.  101 

the  curve  at  P,  and  for  a  similar  reason  SQ  is  tangent  at  Q.  Hence 
the  two  projections  sp  and  sq  are  tangent  to  xpy--qu  at  p  and  q 
respectively;  and  the  vertical  projections  of  the  same  elements 
will  be  tangent  at  p'  and  q' . 

The  plane  of  which  tc'  is  the  horizontal  trace,  is  tangent  to  the 
cone  along  the  element  SK,  and  intersects  the  cylinder  in  two  ele- 
ments, the  horizontal  project!  rs  of  which  are  tangent  to  xp--qv 
at  k  and  v. 

The  projections  of  the  curve  can  now  be  drawn  with  great  accu- 
racy. The  horizontal  projection  of  that  part  which  lies  above  the 
two  elements  SM  and  SL,  and  also  above  the  element  DW,  is 
drawn  full.  Likewise  the  vertical  projection  of  that  part  which 
lies  in  front  of  the  element  BE,  and  also  in  front  of  the  two  ex- 
treme elements  of  the  cone  is  full. 

If  two  auxiliary  planes  be  passed  tangent  to  the  cylinder,  and 
both  intersect  the  cone,  it  is  evident  that  the  cylinder  will  pene- 
trate the  cone,  so  as  to  form  two  distinct  curves  of  intersection. 
If  one  intersects  the  cone  and  the  other  does  not,  a  portion  only 
of  the  cylinder  enters  the  cone,  and  there  will  be  a  continuous 
curve  of  intersection,  as  in  the  figure. 

If  neither  of  these  planes  intersects  the  cone,  and  the  cone  lies 
between  them,  the  cone  will  penetrate  the  cylinder,  making  two 
distinct  curves. 

If  both  planes  are  tangent  to  the  cone,  all  the  rectilinear  ele- 
ments of  both  surfaces  will  be  cut. 


174.  The  intersection  of  two  cylinders  may  be  found,  by  passing 
a  plane  through  a  rectilinear  element  of  one  cylinder  parallel  to 
the  rectilinear  elements  of  the  other,  and  then  intersecting  the 
cylinders  by  a  system  of  planes  parallel  to  this  plane.  The  hori- 
zontal traces  of  these  planes  will  be  parallel,  and  the  construction 
will  be  in  all  respects  similar  to  that  of  the  preceding  problem. 

The  intersection  of  two  cones  may  also  be  found  by  using  a  sys-^ 
tern  of  planes,  through  the  vertices  of  both  cones.     The  right  line, 


102  DESCRIPTIVE   GEOMETRY. 

which  joins  these  vertices,  will  lie  in  all  of  these  planes,  and  pierce 
the  horizontal  plane  in  a  point  common  to  all  the  horizontal  traces. 
We  may  ascertain  whether  the  cylinders  or  cones  intersect  in 
two  distinct  curves,  or  only  one,  in  the  same  manner  as  in  Art. 
(1 73),  hy  passing  auxiliary  planes  tangent  to  either  cylinder  or 
cone. 


1*75.  Problem  45.  To  find  the  intersection  of  a  cone  and  heli- 
coid. 

Let  mnl,  Fig.  72,  be  the  base  of  the  cone,  S  its  vertex ;  and  let 
prq  be  the  horizontal,  and  p'o"q'  the  vertical  projection  of  the 
helical  directrix,  (,<?,  o's')  being  the  axis,  and  the  rectilinear  genera- 
trix being  parallel  to  the  horizontal  plane,  Art.  (92). 

Analysis.  Intersect  the  surfaces  by  a  system  of  auxiliary  planes 
through  the  axis  of  the  helicoid.  These  planes  cut  from  both  sur- 
faces rectilinear  elements,  which  intersect  in  points  of  the  required 
curve. 

Construction.  Draw  sk  as  the  horizontal  trace  of  an  auxiliary 
plane.  This  plane  intersects  the  cone  in  the  element  SK,  and  the 
helicoid  in  an  element  of  which  sg  is  the  horizontal  and  h'g'  the 
vertical  projection.  These  intersect  in  X,  a  point  of  the  required 
curve.  In  the  same  way,  Y,  Z,  and  other  points,  are  determined. 
At  the  points  m'  and  z',  the  vertical  projection  of  the  curve  is  tan- 
gent to  s'm'  and  s'V.  Art.  (160). 

To  draw  a  tangent  to  the  curve  at  X ;  pass  a  plane  tangent  to 
each  of  the  surfaces  at  X,  Arts.  (129)  and  (139) :  leu  is  the  horizon- 
tal trace  of  the  plane  tangent  to  the  cone,  and  tw,  parallel  to  sg, 
that  of  the  plane  tangent  to  the  helicoid,  and  their  intersection 
TJX,  is  the  required  tangent  line. 


176.  Problem  46.  To  find  the  intersection  of  a  cylinder  and 
hemisphere. 

Let  mnl  Fig.  73,  be  the  base  of  the  cylinder  and  MX  a  rectili- 


DESCRIPTIVE   GEOMETRY.  103 

near  element.  Let  ced  be  the  horizontal  and  c'fd'  the  vertical 
projection  of  the  hemisphere. 

Analysis.  Intersect  the  surfaces  by  a  system  of  auxiliary  planes 
j  arallel  to  the  rectilinear  elements  of  the  cylinder  and  perpendicu- 
lar to  the  horizontal  plane.  Each  plane  will  cut  from  the  cylinder 
two  rectilinear  elements,  and  from  the  sphere  a  semicircumference 
the  intersection  of  which  will  be  points  of  the  required  curve. 

Construction.  Take  pn  as  the  trace  of  an  auxiliary  plane.  It 
cuts  from  the  cylinder  the  two  elements  PY  and  NZ,  and  from  the 
hemisphere  a  semicircumference  of  which  i  is  the  centre  and  ig 
the  radius.  Revolve  this  plane  about  pn  until  it  coincides  with  H. 
pi"  will  be  the  revolved  position  of  PY,  Art  (28),  and  nz"  parallel 
topi"  the  revolved  position  of  NZ;  gcz"  the  revolved  position  of 
the  semicircumference;  y"  and  z"  the  revolved  position  of  the  re- 
quired points,  arid  Y  and  Z  the  points  in  their  true  position. 

In  the  same  way  any  number  of  points  may  be  determined. 

The  points  of  tangency  x  and  tv  are  found  by  using  mx  and  Iw 
as  the  traces  of  auxiliary  planes,  and  /  and  u\  by  using  or  and  su* 

A  tangent  at  any  point  may  be  constructed  by  finding  the  inter- 
section of  two  planes  tangent  to  the  cylinder  and  sphere  as  in  Arts, 
(123)  and  (145). 

The  tangents  at  Z  and  Y  are  evidently  parallel  to  H. 


177.  Problem  47.  To  find  the  intersection  of  a  cone  and  hemi- 
sphere* 

Let  mlo,  Fig.  74,  be  the  base  of  the  cone,  and  S  its  vertex,  at 
the  centre  of  the  sphere ;  abc  being  the  horizontal  and  cUd'e'  the 
vertical  projection  of  the  hemisphere. 

Analysis,  Intersect  the  surfaces  by  a  system  of  planes  passing 
through  the  vertex  and  perpendicular  to  the  horizontal  plane. 
Each  plane  will  cut  from  the  cone  two  rectilinear  elements,  and 
from  the  hemisphere  a  semicircumference,  the  intersection  of  which 
will  be  points  of  the  required  curve. 

Construction.  Take  sp  as  the  horizontal  trace  of  one  of  the  auxil 


104  DESCRIPTIVE   GEOMETRY. 

iary  planes.  It  intersects  the  cone  in  the  two  elements  SP  and  SQ, 
and  the  hemisphere  in  a  semicircle  whose  centre  is  at  S.  Revolve 
this  plane  about  the  horizontal  projecting  line  of  S,  until  it  becomes 
parallel  to  V.  The  element  SP  will  be  vertically  projected  in 
a'p'",  SQ  in  s'q'",  the  semicircle  in  a'd'c',  and  x"  and  y"  will  be 
the  vertical  projections  of  the  revolved  positions  of  the  points  of 
intersection.  In  the  counter  revolution  these  points  describe  the 
arcs  of  horizontal  circles  and  in  their  true  position  will  be  verti- 
cally projected  at  x'  and  ?/,  and  horizontally  at  x  and  y. 

In  the  same  way  any  number  of  points  may  be  found. 

The  points  of  tangency  u'  and  z'  are  found  by  using  auxiKary 
planes  which  cut  out  the  extreme  elements  SM  and  SO. 

The  points  in  which  the  vertical  projection  of  the  curve  is  tan- 
gent to  the  semicircle  a'd'c,  are  found  by  using  the  auxiliary  plane 
whose  trace  is  st. 

To  draw  a  tangent  to  the  curve  at  any  point,  we  pass  a  plane 
tangent  to  the  sphere  at  this  point  as  in  Art.  (145)  and  also  one 
tangent  to  the  cone  at  the  same  point  as  in  Art  (129),  and  deter- 
mine their  intersection. 


178.  Problem  48.      To  develop  an  oblique  cone  with  any  base. 

Let  the  cone  be  given  as  in  the  preceding  problem,  Fig.  *74,  and 
let  it  be  developed  on  the  plane  tangent  along  the  element  SP. 

Analysis.  If  the  cone  be  intersected  by  a  sphere  having  its  cen- 
tre at  the  vertex,  all  the  points  of  the  curve  of  intersection,  will  be 
at  a  distance  from  the  vertex  equal  to  the  radius  of  the  sphere ; 
hence,  when  the  cone  is  developed,  this  curve  will  develop  into  the 
arc  of  a  circle  having  its  centre  at  the  position  of  the  vertex,  and 
its  radius  equal  to  that  of  the  sphere.  On  this  we  can  lay  off  the 
rectified  arcs  of  the  curve  of  intersection,  included  between  the 
several  rectilinear  elements,  Art.  (159),  and  then  draw  these  ele- 
ments to  the  position  of  the  vertex. 

The  developed  base,  or  any  curve  on  the  surface,  may  be  traced 
on  the  plane  of  development,  by  laying  off  on  each  element,  from 


DESCRIPTIVE    GEOMKTRY.  105 

the  vertex,  the  distance  from  the  vertex  to  the  point  where  the 
element  intersects  the  base  or  curve.  A  line  through  the  extrem- 
ities of  these  distances  will  be  the  required  development. 

Construction.  Find  as  in  the  preceding  problem  the  curve 
XUY---.  With  S,  Fig.  75,  as  a  centre,  and  sa  as  a  radius,  describe 
the  arc  XUR.  It  is  the  indefinite  development  of  the  intersec- 
tion of  the  sphere  and  cone.  Draw  SX  for  the  position  of  the 
element  SX. 

To  find  the  distance  between  any  two  points  measured  on  the 
curve  XUY---,  we  first  develop  its  horizontal  projecting  cylinder 
on  a  plane  tangent  to  it  at  X,  as  in  Art  (161).  X'U'R'  Fig.  a, 
is  the  development  of  the  curve.  On  XUR  lay  off  XU=X'U', 
UR=U'R',  &c,  and  draw  SU,  SR,  &c.  These  will  be  the  posi- 
tions of  the  elements  on  the  plane  of  development.  On  these  lay 
off  SP=s'jt/",  SM=£*W,  <fec,  and  join  the  points  PM,  &c,  and 
we  have  the  development  of  the  base  of  the  cone. 


1*79.  Problem  49.  To  find  the  intersection  of  two  surfaces  of 
revolution,  whose  axes  are  in  the  same  plane. 

J,  First,  jet  the  axes  intersect  and  let  one  of  the  surfaces  be  an 
ellipsoid  of  revolution  and  the  other  a  paraboloid;  and  let  the 
horizontal  plane  be  taken  perpendicular  to  the  axis  of  the  ellipsoid 
and  the  vertical  plane  parallel  to  the  axes;  (c,c'd'),  Fig.  76,  being 
the  axis  of  the  ellipsoid,  and  (c-l,  s'l')  that  of  the  paraboloid.  Let 
the  ellipsoid  be  represented  as  in  Art.  (107)  and  let  z'f'r'  be  the 
vertical  projection  of  the  paraboloid. 

Analysis.  Intersect  the  two  surfaces  by  a  system  of  auxiliary 
spheres  having  their  centres  at  the  point  of  intersection  of  the 
axes.  Each  sphere  will  intersect  each  surface  in  the  circumference 
of  a  eirele  perpendicular  to  its  axis,  Art.  (97).  and  the  points  of  inter- 
section of  these  circumferences  will  be  points  of  the  required  curve. 

Construction.  With  *'  as  a  centre  and  any  radius,  s'q\  describe 
the  circle  q'p'r' ;  it  will  be  the  vertical  projection  of  an  auxiliary 
sphere.     This  sphere  intersects  the  ellipsoid  in  a  circumference 


106  DESCRIPTIVE    GKOMETKY. 

vertically  projected  in  p'q\  and  horizontally  in  pxq.  It  intersects 
the  paraboloid  in  a  circumference  vertically  projected  in  r'v'. 
These  circumferences  intersect  in  two  points  vertically  projected 
at  x\  and  horizontally  at  x  and  x" '. 

In  the  same  way  any  number  of  points  may  be  found. 

The  points  on  the  greatest  circle  of  the  ellipsoid  are  found  by 
using  s'n'  as  a  radius.  These  points  are  horizontally  projected  at 
u  and  u'\  points  of  tangency  of  xzx"  with  nom. 

The  points  W  and  Z  are  the  points  in  which  the  meridian 
curves  parallel  to  V  intersect,  and  are  points  of  the  required  curve. 

Each  point  of  the  curve  z'x'w'  is  the  vertical  projection  of  two 
points  of  the  curve  of  intersection,  one  in  front  and  the  other  be- 
hind the  plane  of  the  axes. 

A  tangent  may  be  drawn  to  the  curve  at  any  point  as  X,  as  in 
Art.  (1 12).  Otherwise  thus  :  If  to  each  surface  a  normal  line  be 
drawn  at  X,  the  plane  of  these  two  normals  will  be  perpendicular 
to  the  tangent  plane  to  each  surface  at  X,  and  therefore  perpen- 
dicular to  their  intersection,  which  is  the  required  tangent  line. 
Art.  (11 2).  Hence  if  through  X  a  right  line  be  drawn  perpen- 
dicular to  this  normal  plane,  it  will  be  the  required  tangent. 

Since  the  meridian  plane  to  a  surface  of  revolution  is  normal 
to  the  surface,  Art.  (115),  the  normal  to  each  surface  at  X  must 
lie  in  the  meridian  plane  of  the  surface  and  therefore  intersect  the 
axis. 

To  construct  the  normal  to  the  ellipsoid,  revolve  the  meridian 
curve  through  X,  about  the  axis  of  the  surface,  until  it  becomes 
parallel  to  V  ;  it  will  be  projected  into  c'n'd1  and  the  point  X  will 
be  vertically  projected  at  q'.  Perpendicular  to  the  tangent  at  q' 
draw  q'k'.  It  will  be  the  vertical  projection  of  the  normal  in  its 
revolved  position.  After  the  counter-revolution,  K  remaining 
fixed,  k'x'  will  be  the  vertical,  and  ex  the  horizontal  projection  of 
the  normal. 

In  the  same  way  the  normal  LX  to  the  paraboloid  may  be  con- 
structed. 

The  line  k'l'  is  the  vertical  projection  of  the  intersection  of  the 


DESCRIPTIVE    GEOMETRY.  107 

plane  of  these  two  normals  with  the  plane  of  the  axes,  and  is 
parallel  to  the  vertical  trace  of  the  first  plane ;  hence  x't'  perpen- 
dicular to  this  line  is  the  vertical  projection  of  the  required  tan- 
gent, xi  is  the  horizontal  projection  of  the  intersection  of  the  nor- 
mal plane  with  the  plane  of  the  circle  PXQ,  and  this  is  parallel  to 
the  horizontal  trace  of  the  normal  plane ;  hence  xt  perpendicular 
to  this  is  the  horizontal  projection  of  the  required  tangent. 

Second.  If  the  axes  of  the  two  surfaces  are  parallel,  the  construc- 
tion is  more  simple,  as  the  auxiliary  spheres  become  planes  per- 
pendicular to  the  axes,  of  parallel  to  the  horizontal  plane.  Let 
the  construction  be  made  in  this  case. 


PRACTICAL    PROBLEMS. 

180.  In  the  preceding  articles  we  have  all  the  elementary 
principles  and  rules,  relating  to  the  orthographic  projection.  Thft 
student  who  has  thoroughly  mastered  them  will  have  no  diffi- 
"culty  in  their  application. 

Let  this  application  now  be  made  to  the  solution  of  the  follow- 
ing simple  problems. 


181.  Problem  50.  Having  given  two  of  the  faces  of  a  triedral 
angle,  and  the  diedral  angle  opposite  one  of  them,  to  construct  the 
triedral  angle. 

Let  dsf  and  fie",  Fig.  77,  be  the  two  given  faces  and  A  the 
given  angle  opposite /se",  dsf  being  in  the  horizontal  plane,  and 
the  vertical  plane  perpendicular  to  the  edge  sf. 

Construct,  as  in  Art.  (54),  de'  the  vertical  trace  of  a  piano 
whose  horizontal  trace  is  sd  and  making  wi£h  H  the  angle  A  ;  sde' 
will  be  the  true  position  of  the  required  face.  Revolve  fie"  abou 
sf  until  the  point  e"  comes  into  de'  at  e'.  This  must  be  the  poiu 
in  which  the  third  edge  in  true  position  pierces  V  Join  c'f;  i 
will  be  the  vertical  trace  of  the  plane  of  the  face  fie",  in  true  po- 
sition, and  SE  will  be  the  third  edge. 


108  DESCRIPTIVE   GKOMKTET. 

Revolve  sde'  about  sd  until  it  coincides  with  H;  ef  falls  at  e"f 
Art.  (IV),  and  due'"  is  the  true  size  of  the  third  or  required  face. 

e'fd  is  the  diedral  angle  opposite  the  face  dse' ;  and  pvq,  deter- 
mined as  in  Art.  (52),  is  the  diedral  angle  opposite  dsf. 


182.  Problem  51.  Having  given  two  diedral  angles  formed  by 
the  faces  of  a  triedral  angle,  and  the  face  opposite  one  of  them,  to  con- 
struct the  angle. 

Let  A  and  B,  Fig.  78,  be  the  two  diedral  angles,  and  dse'"  the 
face  opposite  B. 

Make  e'df  =  A.  Revolve  dse'"  about  ds  until  e'"  comes  into 
de  at  e'.  This  will  be  the  point  where  the  edge  se" ,  in  true  po- 
sition, pierces  V,  and  SE  will  be  this  edge. 

Draw  e'm  making  e'm.e  =  B,  and  revolve  em  about  e'e',  it  will 
generate  a  right  cone  whose  rectilinear  elements  all  make,  with  H, 
an  angle  equal  to  B.  Through  s  pass  the  plane  sfe'  tangent  to 
this  cone,  Art.  (130).  It,  with  the  faces  fsd  and  sde',  will  form 
the  required  angle. 

fse"  is  the  true  size  of  the  face  opposite  A,  e"  being  the  revolved 
position  of  e' ,  determined  as  in  Art.  (17),  and  the  third  diedral 
angle,  formed  by  e'fs  and  e'ds  may  be  found  as  in  Art  (52). 


183.  Problem  52.  Given  two  faces  of  a  triedral  angle  and 
their  included  diedral  angle,  to  construct  the  angle. 

Let  dse'"  and  dsf,  Fig.  79,  be  the  two  given  faces,  and  A  the 
given  angle.  / 

Make  e'df  equal  A.  de'  will  be  the  vertical  trace  of  the  plane 
of  the  face  dse'",  in  its  true  position.  Revolve  dse'"  about  sd  un- 
til e'"  comes  iuto  de'  at  e' ,  the  point  where  the  edge  se'",  in  true 
position,  pierces  V.  Draw  e'f.  It  is  the  vertical  trace  of  the 
plane  of  the  third  or  required  face,  and  fse"  is  its  true  size. 

eon",  Art.  (53),  is  the  diedral  angle  opposite  e'ds,  and  the  third 
diedral  angle  can  be  found  as  in  Art.  (52). 


DESCRIPTIVE    GKOMETRY.  109 

184.  Problem  53.  Given  one  face  and  the  two  adjacent  dicdral 
angles  of  a  triedral  angle,  to  construct  the  angle. 

Let  dsf  Fig.  *79,  be  the  given  face  and  A  and  B  the  two  adja- 
cent diedral  angles. 

Make  e'df=A.  Construct  as  in  Art.  (54),  fe',  the  vertical 
trace  of  a  plane  sfe',  making  with  H  an  angle  =  B.  SE  will  be 
the  third  edge,  dse'"  and  fse"  the  true  size  of  the  other  faces 
and  the  third  diedral  angle  is  found  as  in  Art.  (52). 


185.  Problem  54.  Given  the  three  faces  of  a  triedral  angle,  to 
construct  tlte  angle. 

Let  dse'",  Fig.  80,  dsf  and  fse",  be  the  three  given  faces,  sd 
and  sf  being  the  two  edges  in  the  horizontal  plane. 

Make  se"  =  se'".  Revolve  the  face/se"  about  fs  ;  e"  describes 
an  arc  whose  plane  is  perpendicular  to  II,  Art.  (17),  of  which  e"e 
is  the  horizontal  and  ee'  the  vertical  trace.  Also  revolve  dse'" 
about  ds ;  e'"  describes  an  arc  in  the  vertical  plane.  These  two 
arcs  intersect  at  e',  the  point  wb^re  the  third  edge  pierces  V  and 
SE  is  this  edge. 

Join  e'd  and  e'f.  These  are  the  vertical  traces  of  the  planes  oi 
the  faces  dse"  and/^r",  in  true  position.  The  diedral  angles  may 
now  be  found  as  in  the  preceding  articles. 


186.  Problem  55.  Given  the  three  diedral  angles  formed  by 
the  faces  of  a  triedral  angle,  to  construct  the  angle. 

Let  A,  B,  and  C,  Fig.  81,  be  the  diedral  angles. 

Make  e'df  =  A.  Draw  ds  perpendicular  to  AB  and  take  e'ds 
as  the  plane  of  one  of  the  faces.  If  we  now  construct  a  plane 
which  shall  make,  with  H  and  e'ds,  angles  respectively  equal  to  B 
and  C,  it,  wTith  these  planes,  will  form  the  required  triedral  angle. 

To  do  this ;  with  d  as  a  centre  and  any  radius  as  dm,  describe 
a  sphere;  mn'q  will  be  its  vertical  projection.  Tangent  to  mu\ 
draw  o'u   making  o'ud  =  B,  and   revolve  it  about  o'd.     It  will 


110  DE3CRIJ  FIVE   GEOMETRY. 

generate  a  cone  whose  vertex  is  o',  tangent  to  the  sphere  and  all 
of  whose  rectilinear  elements  make  with  H  an  angle  equal  to  B. 

Also  tangent  to  mn'q,  draw  p'r'  making  with  de'  an  angle  equal 
to  C,  and  revolve  it  about  p'd.  It  will  generate  a  cone  whose 
vertex  is  p' ,  tangent  to  the  sphere,  and  all  of  whose  rectilinear 
elements  make  with  the  plane  sde'  an  angle  equal  to  C.  If  now 
through  o'  and  p'  a  plane  be  passed  tangent  to  the  sphere,  it  will 
be  tangent  to  both  cones,  and  be  the  plane  of  the  required  third 
face,  p'o'  is  the  vertical  trace  of  this  plane,  and/s  tangent  to  the 
base  uxy  is  the  horizontal  trace,  SE  is  the  third  edge  and  dsf,  dse'" 
and  fse'  the  three  faces  in  true  size. 


187.  By  a  reference  to  Spherical  Trigonometry,  it  will  be  seen 
that  the  preceding  six  problems  are  simple  constructions  of  the 
required  parts  of  a  spherical  triangle,  when  either  three  are  given. 
Thus  in  problem  50,  two  sides  a  and  c,  and  an  angle  A  opposite  one, 
are  given,  and  the  others  constructed.  In  problem  51,  two  angles 
A  and  B,  and  a  side  b,  opposite  one,  are  given,  &c. 


188.  Problem  56.  To  construct  a  triangular  pyramid,  having 
given  the  base  and  the  three  lateral  edges. 

Let  cde,  Fig.  82,  be  the  base  in  the  horizontal  plane,  AB  being 
taken  perpendicular  to  cd ;  and  let  cS,  dS,  and  eS  be  the  three 
edges. 

With  c  as  a  centre  and  cS  as  a  radius,  describe  a  sphere,  inter- 
secting H  in  the  circle  mon.  The  required  vertex  must  be  in  the 
surface  of  this  sphere.  With  d  as  a  centre  and  dS  as  a  radius,  de- 
scribe a  second  sphere  intersecting  H  in  the  circle  qmp.  The  re- 
quired vertex  must  also  be  on  this  surface.  These  two  spheres 
intersect  in  a  circle  of  which  mn  is  the  horizontal,  and  m's'n'  the 
vertical  projection,  Art.  (97).  With  us  a  centre,  and  eS  as  a 
radius,  describe  a  third  sphere,  intersecting  the  second  in  a  circle 
of  which  qp  is  the  horizontal  projection.     These  two  circles  inter- 


DESCRIPTIVE    GEOMETRY.  Ill 

sect  in  a  right  line  perpendicular  to  II  at  s,  and  vertically  projected 
in  r's'.  This  line  intersects  the  first  circle  in  S,  which  must  be  a 
point  common  to  the  three  spheres,  and  therefore  the  vertex  of  the 
required  pyramid.  Join  S  with  c,  d  and  e,  and  we  have  the  lateral 
edges,  in  true  position. 


189.  Problem  57.  To  circumscribe  a  sphere  about  a  triangular 
pyramid. 

Let  ?nno,  Fig.  83,  be  the  base  of  the  pyramid  in  the  horizontal 
plane,  and  S  its  vertex. 

Analysis.  Since  each  edge  of  the  pyramid  must  be  a  chord  of 
the  required  sphere,  if  either  edge  be  bisected  by  a  plane  perpen- 
dicular to  it,  this  plane  will  contain  the  centre  of  the  sphere. 
Hence,  it  three  such  planes  be  constructed  intersecting  in  a  point, 
this  must  be  the  required  centre,  and  the  radius  will  be  the  right 
line  joining  the  centre  with  the  vertex  of  either  triedral  angle. 

Construction.  Bisect  mn  and  no,  by  the  perpendiculars  re  and 
pc.  These  will  be  the  horizontal  traces  of  two  bisecting  and  per- 
pendicular planes.  They  intersect  in  a  right  line  perpendicular  to 
H  at  c.  Through  XJ  ihe  middle  point  of  SO  pass  a  plane  perpen- 
dicular to  it,  Art.  (46).  tTt'  is  this  plane.  It  is  pierced  by  the 
perpendicular  (c,  dc')  at  C,  Art.  (42),  which  is  the  intersection  of 
the  three  bisecting  planes,  and  therefore  the  centre  of  the  required 
sphere.    CO  is  its  radius,  the  true  length  of  which  is  c'o'1 ',  Art.  (29). 

With  c  and  c'  as  centres,  and  with  yo"  as  a  radius,  describe  cir- 
cles. They  will  be,  respectively,  the  horizontal  and  vertical  pro- 
jections of  the  sphere. 


190.  Problem  58.  To  inscribe  a  sphere  in  a  ghen  triangular 
pyramid. 

Let  the  pyramid  S-mno,  Fig.  84,  be  given  as  in  the  preceding 
problem. 

Analysis.  The  centre  of  the  required  sphere  must  be  equally 


112  DKSCKIPTIVK    GKOMET.KY- 

distant  from  the  four  faces  of  the  pyramid,  and  therefore  must  be 
in  a  plane  bisecting  the  diedral  angle  formed  by  either  two  of  its 
faces.  Hence,  if  we  bisect  three  of  the  diedral  angles,  by  planes 
intersecting  in  a  point,  this  point  must  be  the  centre  of  the  re- 
quired sphere,  and  the  radius  will  be  the  distance  from  the  centre 
to  either  face. 

Construction.  Find  the  angle  sps"  made  by  the  face  Son  with  H, 
Art.  (53).  Bisect  this  by  the  line  pu,  and  revolve  the  plane  sps" 
to  its  true  position.  The  line  pu,  in  its  true  position,  and  on  will 
determine  a  plane  bisecting  the  diedral  angle  sps".  In  the  same 
way  determine  the  planes  bisecting  the  diedral  angles  srsf",  sqsn". 
These  planes,  with  the  base  wnw,  form  a  second  pyramid,  the  vertex 
of  which  is  the  intersection  of  the  three  planes,  and  therefore  the 
required  centre. 

Intersect  this  pyramid  by  a  plane  parallel  to  H,  whose  vertical 
trace  is  t'y' .  This  plane  intersects  the  faces  in  lines  parallel  to 
mw,  no,_and  om  respectively.  These  lines  form  a  triangle  whose 
vertices  are  in  the  edges.  To  determine  these  lines,  lay  off  pu=z 
y'y" ,  draw  vz  parallel  to  ps.  z  will  be  the  revolved  position  of  the 
point  in  which  the  parallel  plane  intersects  pu  in  its  true  position. 
z"  is  the  horizontal  projection  of  this  point,  and  z" y  of  the  line 
parallel  to  no.  In  the  same  way  ty  and  tx  are  determined.  Draw 
ox  and  ny ;  they  will  be  the  horizontal  projections  of  two  of  the 
edges.  These  intersect  in  c  the  horizontal  projection  of  the  ver- 
tex, n'y'  is  the  vertical  projection  of  the  edge  which  pierces  H 
at  n,  c'  the  vertical  projection  of  the  centre,  and  c'd'  the  radius. 

With  c  and  c'  as  centres  describe  circles  with  c'd'  as  a  radius. 
They  will  be  the  horizontal  and  vertical  projections  of  the  required 
sphere. 


PART  II. 


SPHERICAL  PROJECTIONS. 


PRELIMINARY    DEHtflTIOffS. 

191.  One  of  the  most  interesting  applications  of  the  principles 
of  Descriptive  Geometry  is  to  the  representation,  upon  a  single 
plane,  of  the  different  circles  of  the  earth's  surface ,  regarded  as  a 
perfect  sphere. 

These  representations  are  Spherical  Projections,  The  plane  of 
projection  which  is  generally  taken  as  that  of  one  of  the  great 
circles  of  the  sphere,  is  the  primitive  plane  ;  and  this  great  circle 
is  the  primitive  circle. 

The  axis  of  the  earth  is  the  right  line  about  which  the  earth  is 
known  daily  to  revolve. 

The  two  points  in  which  it  pierces  the  surface  are  the  poles,  one 
being  taken  as  the  North  and  the  other  as  the  South  Pole. 

The  axis  of  a  circle  of  the  sphere  is  the  right  line  through  its 
centre  perpendicular  to  its  plane,  and  the  points  in  which  it  pierces 
the  surface  are  the  poles  of  the  circle. 

The  polar  distance  of  a  point  of  the  sphere  is  its  distance  from 
either  pole  of  the  primitive  circle. 

The  polar  distance  of  a  circle  of  the  sphere  is  the  distance  of  any 
point  of  its  circumference  from  either  of  its  poles. 

8 


114  DESCRIPTIVE   GEOMETRY. 

192.  The  lines  on  the  earth's  surface,  usually  represented,  are : 

1.  The  Equator,  the  circumference  of  a  great  circle  whose  plane 
is  perpendicular  to  the  axis. 

2.  The  Ecliptic,  the  circumference  of  a  great  circle  making  an 
angle  of  23^-°,  nearly,  with  the  equator.  It  intersects  the  equator 
in  two  points,  called  the  Equinoctial  Points. 

3.  The  Meridians,  the  circumferences  of  great  circles  whose 
planes  pass  through  the  axis ;  and  are  therefore  perpendicular  to 
the  plane  of  the  equator. 

Of  these  meridians  two  are  distinguished  I  the  Equinoctial  Co- 
lure,  which  passes  through  the  equinoctial  points ;  and  the  Sols- 
titial Colure^  whose  plane  is  perpendicular  to  that  of  the  equinoc 
tial  col  ure. 

The  solstitial  colure  intersects  the  ecliptic  in  two  points,  called 
the  Solstitial  Points. 

4.  The  Parallels  of  Latitude^  the  circumferences  of  small  circles 
parallel  to  the  equator. 

Four  of  these  are  distinguished  : 

The  Arctic  Circle,  23j°  from  the  north  pole ; 
The  Antarctic  Circle,  23^°  from  the  south  pole 
The  Tropic  of  Cancer,  231°  north  of  the  equator; 
The  Tropic  of  Capricorn,  23^-°  south  of  the  equator. 

The  first  two  are  also  called  Polar  Circles*  \ 


193.  The  Latitude  of  a  point  on  the  earth's  surface,  is  its  dis- 
tance from  the  equator,  measured  on  a  meridian  passing  through 
the  point. 

The  Horizon  of  a  point  or  place,  on  the  earth's  surface,  is  the 
circumference  of  a  great  circle  whose  plane  is  perpendicular  to  the 
radius  passing  through  the  point. 

Let  M,  Fig.  85,  be  any  point  on  the  earth's  surface.     Through 
this  point  and  the  axis  pass  a  plane,  and  let  MNS  be  the  circum 
ference  cut  from  the  sphere ;  N  the  north  and  S  the  south  pole 
ECE'  the  intersection  of  the  plane  with  the  equator,  and  PCQ 


SPHERICAL    PROJECTIONS.  115 

perpendicular  to  CM,  its  intersection  with  the  horizon  of  the  given 
point.  Then  ME  is  the  latitude  of  the  point,  and  NQ  the  distance 
of  the  pole  N  from  the  horizon.  NQ  also  measures  the  angle 
NCQ,  the  inclination  of  the  axis  to  the  horizon.     But 

NQ  =5  ME, 

since  each  is  obtained  by  subtracting  NM  from  a  quadrant;  that 
is,  the  distance  from  either  pole  of  the  earth  to  the  horizon  of  a  place, 
is  equal  to  the  latitude  of  that  place \ 


194.  Let  NS,  Fig.  85,  and  MR  be  the  axes  of  two  circles  inter- 
secting the  plane  NCM  in  the  lines  EE'  and  PQ  respectively. 
ECP  is  then  the  angle  made  by  the  planes  of  these  circles.     But 

ECP  as  NCM, 

since  each  is  obtained  by  subtracting  MCE  from  a  right  angle ;  that 
is,  the  angle  between  any  two  circles  of  the  sphere,  is  equal  to  the 
angle  formed  by  their  axes. 


195.  If  a  plane  be  passed  through  the  axes  of  any  circle  of  the 
sphere  and  of  the  primitive  circle,  its  intersection  with  the  primi- 
tive plane  is  the  line  of  measures  of  the  given  circle.  This  auxiliary 
plane  is  perpendicular  to  the  planes  of  both  circles,  and  therefore 
to  their  intersection ;  hence  the  line  of  measures,  a  line  of  this 
plane,  must  be  perpendicular  to  the  intersection  of  the  given  with 
the  primitive  circle,  and  must  also  pass  through  the  centre  of  the 
primitive  circle.  Thus,  if  EE',  Fig.  85,  is  the  intersection  of  a 
circle  with  the  primitive  plane  NESE',  NS  is  its  line  of  measures. 
Also  NS  is  the  line  of  measures  of  any  small  circle  whose  inter- 
section with  the  primitive  plane  is  parallel  to  EE'. 


116  DESCRIPTIVE    GEOMETRY. 


ORTHOGRAPHIC    PROJECTIONS    OF    THE    SPHERE. 

196.  When  the  point  of  sight  is  taken  in  the  axis  of  the  primi- 
tive circle,  and  at  an  infinite  distance  from  this  circle,  the  projec- 
tions of  the  sphere  are  orthographic,  Art.  (2). 

If  E,  Fig.  85,  be  any  point,  e  will  be  its  orthographic  projection 
on  the  plane  of  a  circle  whose  axis  is  CM.     Blit 

that  is,  the  orthographic  orojection  of  any  point  of  the  surface  of 
a  sphere  is  at  a  distance  from  the  centre  of  the  primitive  circle  equal 
to  the  sine  of  its  polar  distance. 


197.  The  circumference  of  a  circle,  oblique  to  the  primitive 
plane,  is  projected  into  an  ellipse.  For  the  projecting  lines  of  its 
different  points  form  the  surface  of  a  cylinder  whose  intersection 
with  the  primitive  plane  is  its  projection,  Art.  (74),  and  this  inter- 
section is  an  ellipse,  Art.  (160). 

If  the  plane  of  the  circumference  be  perpendicular  to  the  primi- 
tive plane,  its  projection  is  a  right  line,  Art.  (62). 

If  the  plane  of  the  circumference  be  parallel  to  the  primitive 
plane,  its  projection  is  an  equal  circumference,  Art.  (62). 

The  projection  of  every  diameter  of  the  circle  which  is  oblique 
to  the  primitive  plane,  will  be  a  right  line  less  than  this  diameter, 
Art.  (29),  while  the  projection  of  that  one  which  is  parallel  to  the 
primitive  plane,  will  be  equal  to  itself,  Art.  (14).  This  projection 
will  then  be  longer  than  any  other  right  line  which  can  be  drawn 
in  the  ellipse,  and  is  therefore  its  transverse  axis.  Analyt.  Geo., 
Art.  (127). 

The  projection  of  that  diameter  which  is  perpendicular  to  the 
one  which  is  parallel  to  the  primitive  plane,  will  be  perpendicular 


SPHKRICAL    PROJECTIONS.  11  T 

to  this  transverse  axis,  Art.  (36),  and  pass  through  the  centre,  and 
therefore  be  the  conjugate  axis  of  the  ellipse,  Art.  (59). 

This  last  diameter  is  perpendicular  to  the  intersection  of  the 
plane  of  the  given  circle  and  the  primitive  plane,  and  therefore 
makes  with  the  primitive  plane  the  same  angle  as  the  circle ;  and 
one-half  its  projection,  or  the  semi-conjugate  axis  of  the  ellipse,  is 
evidently  the  cosine  of  this  inclination,  computed  to  the  radius  of 
the  given  circle.  Hence,  to  project  any  circle  orthographically, 
we  have  simply  to  find  the  projection  of  that  diameter  which  is 
parallel  to  the  primitive  plane,  and  through  its  middle  point  draw 
a  right  line  perpendicular  to  it,  and  make  it  equal  to  the  cosine  of 
the  angle  made  by  the  circle  with  the  primitive  plane.  The  first 
line  is  the  transverse,  and  the  second  the  semi-conjugate  axis  of 
the  required  ellipse,  which  may  then  be  accurately  constructed  as 
in  Art.  (59). 

It  should  be  remarked  that  the  conjugate  axis  of  the  ellipse 
always  lies  on  the  line  of  measures  of  the  circle  to  be  projected, 
Art.  (195). 


198.  The  line  of  measures  of  a  circle  evidently  contains  the 
projections  of  both  poles  of  the  circle,  Art.  (195)  ;  and  since  the 
arc  which  measures  the  distance  of  either  pole  from  the  pole  of 
the  primitive  circle,  measures  also  the  inclination  of  the  two  cir- 
cles, Art.  (194),  it  follows  that  either  pole  of  a  circle  is  ortho- 
graphically projected  in  its  line  of  measures,  at  a  dista  jce  from  the 
centre  of  the  primitive  circle  equal  to  the  sine  of  its  inclination,  Art. 
(196). 


199.  Problem  59.  To  project  the  sphere  upon  the  plane  of  any 
one  of  its  great  circles. 

Let  E/?E'<7,  Fig.  86,  be  the  primitive  circle  intersecting  the 
equator  in  the  points  E  and  E',  and  making  with  it  an  angle  de-. 
noted  by  A.     Let  E  and  E'  also  be  assumed  as  the  equinoctial 


118  '       DESCRIPTIVE   GEOMETRY. 

points.  The  line  EE'  is  then  the  intersection  of  the  primitive 
plane  by  the  equator,  ecliptic,  and  equinoctial  colure,  and  pq  per- 
pendicular to  it  is  the  line  of  measures  of  all  these  circles. 

Let  us  first  project  the  hemisphere  lying  between  the  primitive 
plane  and  the  north  pole. 

Since  EE'  is  that  diameter  of  the  equator  which  lies  in  the 
primitive  plane,  it  is  its  own  projection,  and  therefore  the  trans- 
verse axis  of  the  ellipse  into  which  the  equator  is  projected. 
From  q  lay  off  qm'  =  A,  and  draw  m'm  perpendicular  to  pq. 
Cm  =  cos  A,  and  is  the  semi-conjugate  axis,  Art.  (197).  On  this 
and  EE'  describe  the  semi-ellipse  EmE' ;  it  is  the  projection  of 
that  part  of  the  equator  lying  above  the  primitive  plane. 

n  is  the  projection  of  the  north  pole,  En'  being  made  equal  to 
A,  and  Gn  =  n'x  its  sine,  Art.  (198). 

EE'  is  also  the  transverse  axis  of  the  projection  of  the  ecliptic. 
If  the  portion  of  the  ecliptic  on  the  hemisphere  under  considera- 
tion, lies  between  the  equator  and  the  north  pole,  it  will  make  an 
angle  with  the  primitive  plane  greater  than  that  of  the  equator  by 
23^°.  If  it  lies  between  the  equator  and  the  south  pole,  it  will 
make  a  less  angle  by  '231°.  Taking  the  former  case,  lay  off 
m'o'  =  231°,  then  qo'  =  A  +  23*-°,  and  Co  =  cos  (A  +  23J°)  = 
the  semi-conjugate  axis,  with  which  and  EE'  describe  the  semi- 
ellipse  EoE',  the  projection  of  one  half  the  ecliptic. 

The  equinoctial  colure,  making  with  the  equator  a  right  angle, 
makes  with  the  primitive  plane  an  angle  equal  to  90°  +  A.  EE' 
is  the  transverse  axis  of  its  projection,  and  Cn  =  cos  qn'  =  cos 
(90°  +  A)  =  the  semi- conjugate  axis.  And  the  semi-ellipse  EnE' 
is  the  projection  of  that  half  above  the  primitive  plane. 

The  solstitial  colure  being  perpendicular  to  the  equator  and 
equinoctial  colure  is  perpendicular  to  EE',  and  therefore  to  the 
primitive  plane;    hence  its  projection  is  the  right  line  qp,  Art.: 
(197). 

To  project  any  meridian,  as  that  which  makes  with  the  solsti- 
tial colure,  an  angle  denoted  by  B ;  pass  a  plane  tangent  to  the 
sphere  at  the  north  pole.     It  will  intersect  the  planes  of  the  given 


SPHERICAL   PROJECTIONS.  119 

meridian  and  solstitial  colure  in  lines  perpendicular  to  the  axis, 
and  making  with  each  other  an  angle  equal  to  B  ;  and  these  lines 
will  pierce  the  primitive  plane  in  points  of  the  intersections  of  the 
planes  of  these  meridians  with  the  primitive  plane,  Art.  (30).  To 
determine  this  tangent  plane ;  revolve  the  solstitial  colure  about 
pq,  as  an  axis,  .until  it  comes  into  the  primitive  plane.  It  will 
then  coincide  with  l&i'pq,  and  the  north  pole  will  fall  at  n'. 
Draw  n't  tangent  to  Ym'p ;  it  is  the  revolved  position  of  the  in- 
tersection of  the  required  tangent  plane  by  the  plane  of  the 
solstitial  colure.  It  pierces  the  primitive  plane  at  t,  and  st  perpen- 
dicular to  Cw,  Art  (145),  is  the  trace  of  the  tangent  plane.  Re- 
volve this  plane  about  ts  until  it  coincides  with  the  primitive  plane. 
The  north  pole  falls  at  n" ';  tn"  being  equal  to  tn\  Art.  (17). 
Through  n'\  draw  n"s,  making  with  n"t  an  angle  equal  to  R 
This  will  be  the  revolved  position  of  the  intersection  of  the  tan- 
gent plane  by  the  plane  of  the  given  meridian*  It  pierces  the 
primitive  plane  at  s,  and  &C  is  the  intersection  of  the  meridian 
plane  with  the  primitive,  and  yz  is  the  transverse  axis  of  the  re- 
quired projection^  Art.  (197). 

To  find  the  semi-conjugate  ;  through  the  north  pole  pass  a  plane 
perpendicular  to  yz ;  nv  is  its  trace.  Revolve  this  plane  about  nv 
until  it  coincides  with  the  primitive  plane.  N  falls  at  ny",  and 
n'"vn  is  the  angle  made  by  the  meridian  with  the  primitive  plane, 
Art.  (53).  Lay  off  vk  —  CE,  and  draw  ku  parallel  to  yz.  Cu 
is  the  required  semi-conjugate  axis,  and  the  semi-ellipse  yuz  is 
the  projection  of  that  half  of  the  meridian  which  lies  above  the 
primitive  plane. 

Since  the  plane  of  any  parallel  of  latitude,  as  the  arctic  circle, 
is  parallel  to  the  equator,  it  will  be  intersected  by  the  plane  of  the 
equinoctial  colure  in  a  diameter  parallel  to  EEr,  and  to  the  prim- 
itive plane,  and  the  projection  of  this  diameter  will  be  the  trans- 
verse axis  of  the  projection.  To  determine  it;  revolve  the 
equinoctial  colure  about  EE'  as  an  axis  until  it  coincides  with 
EpE'q  ;  N  falls  at  p.  From  p  lay  off  pa'  =  23^°,  the  polar  dis 
tance  of  the  parallel,  and  draw  a'b\     It  will  be  the  revolved  posi 


120  DESCRIPTIVE    GEOMETRY. 

tion  of  that  diameter  of  the  arctic  circle  which  is  parallel  to  the 
primitive  plane.  When  the  colure  is  revolved  to  its  true  position 
a'b'  will  be  projected  into  ab,  the  required  transverse  axis.  From 
d  its  middle  point  lay  off  di  =  cos  A,  computed  to  the  radius  da ; 
it  will  be  the  semi-conjugate  axis,  and  the  ellipse  aibh  is  the  re- 
quired projection. 

In  the  same  way  the  tropic  of  Cancer  or  any  other  parallel  may 
be  projected. 

If  the  polar  distance  of  the  parallel  is  greater  than  90°  —  A,  the 
inclination  of  the  axis,  the  parallel  will  pass  below  the  primitive 
plane  and  a  part  of  its  projection  be  drawn  broken. 

The  projection  of  the  tropic  of  Cancer  is  tangent  to  EoE'  at  of 
Art.  (65). 


200.  Each  point  in  the  primitive  circle  is  evidently  the  projec- 
tion of  two  points  of  the  surface  of  the  sphere,  one  above  and 
the  other  below  the  primitive  plane.  To  represent  these  points 
distinctly  and  prevent  the  confusion  of  the  drawing,  we  first  pro- 
ject the  upper  hemisphere,  as  above,  and  then  revolve  the  lower 
180°,  about  a  tangent  to  the  primitive  circle  at  E.  It  will  then 
be  above  the  primitive  plane  and  may  be  projected  in  the  same 
way  as  the  first.  Ew"E"  is  the  projection  of  the  other  half  of 
the  equator ;  s  of  the  south  pole ;  Eo"E"  of  the  other  half  of  the 
ecliptic;  EsE"  of  the  equinoctial  colure;  y'sz'  of  the  meridian,  &c. 

201.  If  the  projection  be  made  on  the  equator,  the  preceding 
problem  is  much  simplified.  Thus,,  let  EpE'q,  Fig.  87,  be  the 
equator,  n  is  the  projection  of  the  north  pole ;  EoE'  of  one  half 
of  the  ecliptic,  go'  being  equal  to  231°,  and  no  its  cosine. 

Since  the  meridians  are  all  perpendicular  to  the  equator,  EE'  is 
the  projection  of  the  equinoctial,  and  pq  of  the  solstitial  colure; 
yz  of  the  meridian  making  an  angle  of  30°  with  the  solstitial 
colure. 

Since  the  parallels  of  latitude  are  parallel  to  the  primitive  plane, 


SPHERICAL   PROJECTIONS.  121 

altbi  is  the  projection  of  the  arctic  circle,  and  ogkl  that  of  the 
tropic  of  Cancer,  na  being  equal  to  the  sine  of  23^°,  and  nl  = 
sin  06i°,  Art  (196).     ogkl  is  tangent  to  EoE'  at  o. 


202.  If  the  projection  be  made  on  the  equinoctial  colure  ;  let 
ENE'S,  Fig.  88,  be  the  primitive  circle,  and  E  and  E'  the  equi- 
noctial points. 

Since  the  equator  is  perpendicular  to  the  primitive  plane,  EEf 
will  be  its  projection.  N  is  the  north  and  S  the  south  pole. 
EoE'  is  the  projection  of  one  half  the  ecliptic;  Co  being  equal  to 
cos  66|-0.     NS  is  the  projection  of  the  solstitial  colure. 

Since  the  parallels  are  perpendicular  to  the  primitive  plane,  ab 
and  a'b'  are  the  projections  of  the  polar  circles ;  Na  and  Sa'  being 
each  equal  to  231°  ;  and  Ig  and  Vg*  the  projections  of  the  tropics. 
N#  and  Sg'  being  each  equal  to  66^°. 

NyS  is  the  projection  of  one  half  the  meridian,  making  an  angle 
of  30°  with  the  solstitial  colure,  or  60°  with  the  primitive  plane, 
Cy  being  equal  to  cos  60°  =  ^  CE'. 


203.  The  projections  may  be  made  upon  the  ecliptic,  and 
horizon  of  a  place,  in  the  same  way  as  in  problem  59.  In  the 
former  case,  the  angle  A  will  equal  23ie ;  and  in  the  latter,  since 
the  angle  included  between  the  axis  and  horizon  is  equal  to  the 
latitude  of  the  place,  Art.  (193),  the  angle  A  between  the  equator 
and  horizon  will  be  90°  +  the  latitude. 


STEREOORAPHIC    PROJECTIONS    OF    THE    SPHERE. 

204.  The  natural  appearance  and  beauty  of  a  scene-graphic 
drawing  will  depend,  very  much,  upon  the  position  chosen  by  the 
draughtsman,  or  artist,  for  the  point  of  sight  This  should  be  so 
selected  that  a  person  taking  the  drawing  into  his  hand  £>r  exam- 


122  DESCRIPTIVE    GEOMETRY. 

ination,  will  naturally  place  his  eye  at  this  point.  From  any  other 
position  of  the  eye,  the  drawing  will  appear  to  some  extent  dis- 
torted. Hence  it  is,  that  an  orthographic  drawing  never  appears 
perfectly  natural,  as  it  is  impossible  to  place  the  eye  of  the  ob- 
server at  an  infinite  distance  from  it. 

In  spherical  projections,  if  the  point  of  sight  be  taken  at  either 
pole  of  the  primitive  circle,  the  projections  are  Stereographic,  and, 
in  general,  present  the  best  appearance  to  the  eye  of  an  ordinary 
observer,  as,  in  this  case,  the  projections  of  all  circles  of  the  sphere, 
as  will  be  seen  in  Art.  (207),  are  circles. 


205.  The  projection  of  each  point  on  the  surface  of  the  sphere, 
will  be  that  point  in  which  a  right  line,  through  it  and  the  point 
of  sight,  pierces  the  primitive  plane,  Art.  (3). 

Let  M,  Fig.  89,  be  any  point  on  the  surface  of  the  sphere. 
Through  it  and  the  axis  of  the  primitive  circle  pass  a  plane.  It 
will  intersect  the  sphere  in  a  great  circle  EME'S,  and  the  prim- 
itive plane  in  a  right  line  EE' ;  N  and  S  being  the  poles  of  the 
primitive  circle  and  S  the  point  of  sight.  NM  is  the  polar  dis- 
tance and  m  the  stereographic  projection  of  M.  Cm  is  the  tan- 
gent of  the  arc  Co,  computed  to  the  radius  CS*=CE,  and  Co  is 
one  half  of  NM.  That  is,  the  stereographic  projection  of  any 
point  of  the  surface  of  the  sphere  is  at  a  distance  from  the  centre 
of  the  primitive  circle  equal  to  the  tangent  of  one  half  its  polar  dis* 
tance. 

In  this  projection,  it  should  be  observed  that  the  polar  distance 
of  a  point  is  always  its  distance  from  the  pole  opposite  the  point 
of  sight,  and  often  exceeds  90Q. 


206.  If  a  plane  be  passed  through  the  vertex  of  an  oblique  cone 
with  a  circular  base,  perpendicular  to  this  base  and  through  its 
centre,  such  plane  is  a  principal  plane,  and  evidently  bisects  all 
chords  of  the  cone  drawn  perpendicular  to  it. 


6PHKRICAL   PROJECTIONS.  123 

Let  SAB,  Fig.  90,  be  such  a  plane  intersecting  the  cone  in  the 
elements  SA  and  SB,  and  the  base  in  the  diameter  AB.  If 
this  cone  be  now  intersected  by  a  plane  tTt\  perpendicular  to  the 
principal  plane,  and  making  with  one  of  the  principal  elements,  as 
SA,  an  angle  Sba  equal  to  the  angle  SBA,  which  the  other  makes 
with  the  plane  of  the  base,  the  section  is  a  sub  contrary  section 
and  will  be  the  circumference  of  a  circle.  For,  through  o,  any 
point  of  ba,  which  is  the  orthographic  projection  of  the  curve  of 
intersection  on  the  principal  plane,  pass  a  plane  parallel  to  the 
base.  It  cuts  from  the  cone  the  circumference  of  a  circle,  and  in- 
tersects the  plane  of  the  sub  contrary  section  in  a  right  line  per- 
pendicular to  SAB  at  o,  and  the  two  curves  have,  at  this  point,  a 
common  ordinate.  The  similar  triangles  aod  and  cob  give  the 
proportion 

ao  :  oc  : :  od  :  ob  ;      or,      ao  X  ob  =■  oc  X  od. 

But  oc  X  od  is  equal  to  the  square  of  the  common  ordinate, 
since  the  parallel  curve  is  a  circle;  hence  ao  X  ob  is  equal  to  the 
square  of  the.  ordinate  of  the  sub  contrary  section,  which  must, 
therefore,  be  a  circle. 


207.  To  project  any  circle  of  the  sphere ;  through  its  axis  and 
the  axis  of  the  primitive  circle  pass  a  plane,  and  let  ENE'S,  Fig. 
89,  be  the  circle  cut  from  the  sphere  by  this  plane  ;  S  the  point 
of  sight;  RM  the  orthographic  projection  of  the  given  circle  on 
the  cutting  plane,  Art.  (62) ;  CN  the  axis  of  the  primitive  circle 
orthographically  projected  in  EE',  and  CI*  the  axis  of  the  given 
circle. 

The  projecting  lines,  drawn  from  points  of  the  circumference  t( 
S,  form  a  cone  whose  intersection  by  the  primitive  plane  will  evi . 
dently  be  the  stereographic  projection  of  the  circumference 
SUM  is  the  principal  plane  of  this  cone,  and  SR  and  SM  the  prin 
cipal   elements.      The   primitive    plane   is  perpendicular   to    this 


124  DESCRIPTIVE    GEOMETRY. 

plane,  and  intersects  the  cone  in  a  curve  of  which  rm  is  the  or- 
thographic projection.     But  the  angle 

Si-m  =  SMR. 

Since  each  is  measured  by  SR,  ES  being  equal  to  E'S,  hence 
this  section  is  a  sub  contrary  section,  and  therefore  a  circle  whose 
diameter  is  mr.  That  is,  the  stereographic  projection  of  every  circle 
on  the  surface  of  a  sphere,  whose  plane  does  not  pass  through  the 
point  of  sight,  is  a  circle. 

mr  is  also  the  line  of  measures  of  the  given  circle,  Art.  (195), 
and  evidently  contains  the  centre  of  its  projection. 

The  distance 

O  =  tan  Co'  =  tan  J  (PR  +  PN), 
and 

Cm  =  tan  Co  =  tan  \  (PR  -  PN). 

Hence,  the  extremities  of  a  diameter  of  the  projection  of  any 
circle,  on  the  surface  of  the  sphere,  are  in  its  line  of  measures,  one 
at  a  distance  from  the  centre  of  the .  primitive  circle,  equal  to  the 
tangent  of  one-half  the  sum  of  the  polar  distance  and  inclination  ol 
the  circle,  and  the  other  at  a  distance  equal  to  the  tangent  of  one- 
half  the  difference  of  these  two  arcs. 

When  the  polar  distance  is  greater  than  the  inclination,  these 
extremities  will  evidently  be  on  different  sides  of  the  centre  of  the 
primitive  circle.  When  less,  they  will  be  on  the  same  side.  If 
the  polar  distance  is  equal  to  the  inclination,  the  projection  of  the 
given  circle  will  pass  through  the  centre  of  the  primitive  circle. 

The  polar  distance  and  inclination  of  any  circle  being  known,  a 
diameter  of  its  projection  can  thus  be  constructed,  and  thence  the 
projection. 


208.  If  the  circle  be  paralle1  to  the  primitive  plane,  the  ?nh 


SPI1KKICAL    PKOJEOTIONS.  125 

contran  and  parallel  sections  coincide,  and  the  projection  is  a  cir- 
cle whose  centre  is  at  the  centre  of  the  primitive  circle,  and  radius 
the  distance  of  the  projection  of  any  point  of  the  circumference 
from  the  centre  of  the  primitive  circle;  that  is,  the  tangent  of  half 
the  circle's  polar  distance,  Art.  (205). 

If  the  plane  of  the  circle  pass  through  the  point  of  sight,  the 
projecting  cone  becomes  a  plane,  and  the  projection  is  a  right  line. 


209.  If  a  right  line  be  tangent  to  a  circle  of  the  sphere,  its  pro- 
jection will  be  tangent  to  the  projection  of  the  circle.  For,  the  pro- 
jecting lines  of  the  circumference  form  a  cone,  and  those  of  the 
tangent,  a  plane  tangent  to  this  cone,  along  the  projecting  line  of 
the  point  of  contact;  hence  the  intersections  of  the  cone  and  plane 
by  the  primitive  plane,  are  tangent  to  each  other  at  the  projection 
of  the  point  of  contact,  Art.  (112).  But  the  first  is  the  projection 
of  the  circle,  and  the  second  that  of  the  tangent. 


210.  Let  MR  and  MT,  Fig.  91,  be  the  tangents  to  two  circles  of 
the  sphere  at  a  common  point,  M.  Let  these  tangents  be  projected 
on  the  primitive  plane  by  the  planes  RMS  and  TMS  respectively, 
in  the  lines  mr  and  mt,  and  let  M«S  and  M6S  be  the  circles  cut 
from  the  sphere  by  these  planes,  and  let  SR  and  ST  be  the  lines 
cut  from  the  tangent  plane  to  the  sphere  at  S.  Since  this'tangent 
plane  is  parallel  to  the  primitive  plane,  the  lines  SR  and  ST  will 
be  parallel  respectively  to  mr  and  mi,  and  the  angle  RST  =  rmt. 
Join  RT.  Since  RS  and  EM  are  each  tangent  to  the  circle  MaS, 
they  are  equal,  and  for  the  same  reason  TM  =  TS ;  hence  the  two 
triangles,  RMT  and  RST,  are  equal,  and  the  angle 

RMT  =  RST  ==  rmt, 

that  is,  the  angle  between  any  two  tangents  to  circles  of  the  sphere 
at  a  common  point,  is  equal  to  the  angle  of  their  projections. 

The  angle  between  the  circles  is  the  same  as  that  between  thei 


126  DESCRIPTIVE   GEOMETRY. 

tangents,  and  since  the  projections  of  the  tangents  are  tangent  to 
the  projections  of  the  circles,  the  angle  between  the  projections  of 
the  circles  is  the  same  as  that  between  the  projections  of  the  tan- 
gents ;  hence  the  angle  between  any  two  circumferences  or  arcs  is 
equal  to  the  angle  between  their  projections. 


211.  If  from  the  centres  of  the  projections  of  two  circles  radii  be 
drawn  to  the  intersection  of  these  projections,  they  will  make  the 
same  angle  as  the  circles  in  space.  For,  these  radii  being  perpen- 
dicular to  the  tangents  to  the  projections,  at  their  common  point, 
make  the  same  angle  as  these  tangents,  and  therefore  as  the  pro- 
jections of  the  arcs,  or  as  the  arcs  themselves. 


212.  If  the  circle  to  be  projected  be  a  great  circle,  it  will  inter- 
sect the  primitive  circle  in  a  diameter  perpendicular  to  its  line  of 
measures,  Art.  (195).  Let  O,  Fig.  92,  be  the  centre  of  the  projec- 
tion of  such  a  circle  intersecting  the  primitive  circle  EPE'R  in  the 
diameter  PR,  CE  being  its  line  of  measures,  and  P  and  R  evident- 
ly points  of  the  projection.  Draw  the  radius  OR.  The  primitive 
circle  is  its  own  projection  ;  therefore  the  angle  CRO  is  equal  to 
the  angle  between  the  given  and  primitive  circles,  Art.  (211).  CO 
is  the  tangent  of  this  angle,  and  OR  its  secant.  Hence  the  centre 
of  the  projection  of  a  great  circle,  is  in  its  line  of  measures,  Art. 
(207),  at  a  distance  from  the  centre  of  the  primitive  circle  equal 
to  the  tangent  of  its  inclination,  and  the  radius  of  the  projection  is 
the  secant  of  this  angle. 


213.  Let  0,  Fig.  93,  be  the  centre  of  the  projection  of  a  small 
circle  perpendicular  to  the  primitive  plane,  and  intersecting  it  in 
PR.  OC  is  its  line  of  measures,  and  P  and  R  points  of  tl-e  pro- 
jection. Join  CR  and  OR.  OR  is  perpendicular  tor  CR,  since 
EPE'R  is  the  projection  of  the  primitive  circle,  and  P/>R,  that  oi 


SPHERICAL    PROJKCTIONS.  127 

the  given  circle,  at  right  angles  with  it,  Art.  (211).  OR  is  there- 
fore the  tangent  of  the  arc  ER,  the  polar  distance  of  the  given 
circle,  and  CO  is  its  secant.  Hence  the  centre  of  the  projection  of 
a  small  circle,  perpendicular  to  the  primitive  plane,  is  in  its  line  of 
measures,  at  a  distance  from  the  centre  of  the  primitive  circle 
equal  to  the  secant  of  the  polar  distance,  and  the  radius  of  the  pro- 
jection is  the  tangent  of  the  polar  distance. 


214.  Let  P  and  R,  Fig.  94,  be  the  poles  of  any  circle  of  the 
sphere ;  EPE'S  being  the  circle  cut  from  the  sphere  by  the  plane 
of  the  axes  of  the  given  and  primitive  circles,  and  MQ,  the  inter- 
section of  this  plane  with  that  of  the  given  circle,  and  EE',  its 
intersection  with  the  primitive  circle,  the  line  of  measures  of  the 
given  circle,  p  is  the  projection  of  P  and  r  of  R.  Qp  is  the  tangent 
of  Co,  equal  to  one-half  of  NP,  which  measures  the  inclination  of 
the  given  circle  to  the  primitive,  Art.  (194).  C/-  is  the  tangent 
of  Co',  the  complement  of  Co,  or  is  the  co-tangent  of  Co.  Hence 
the  poles  of  any  circle  of  the  sphere  are  projected  into  the  line  of 
measures,  the  one  furthest  from  the  point  of  sight  at  a  distance 
from  the  centre  of  the  primitive  circle  equal  to  the  tangent  of  half 
the  inclination,  and  the  other  at  a  distance  equal  to  the  co-tangent 
of  half  the  inclination  of  the  given  to  the  primitive  circle. 


215.  Problem  60.  To  project  the  sphere  upon  the  plane  of  any 
of  its  great  circles,  as  the  ecliptic. 

Let  E/xE'<7,  Fig.  95,  be  the  primitive  circle  intersecting  the 
equator  in  EE7.  In  this  case,  E  and  E'  will  be  the  equinoctial 
points,  and  in  any  other  case  may  be  taken  as  such.  EK'  will 
also  be  the  intersection  of  the  plane  of  the  equinoctial  colure  with 
the  primitive  plane,  and  pq  the  line  of  measures  of  both  these 
circles. 

We  will  first  project  the  hemisphere  above  the  primitive  plane, 
the  point  of  sight  being  at  the  pole  underneath. 


128  DESCRIPTIVE    GEOMETRY. 

Since  the  Equator  makes  an  angle  of  23J°  with  the  primitive 
plane,  we  draw  E'o,  making  the  angle  CE'o  =  231°.  Co  is  the 
tangent  of  this  angle  and  E'o  the  secant;  hence  with  o  as  a  centre 
and  E'o  as  a  radius,  describe  the  arc  EmE' ;  it  is  the  projection  of 
the  part  of  the  equator  above  the  primitive  plane,  Art.  (212). 

From  E  lay  off  Ek  —  231°  and  draw  hE' ;  Cn  is  the  tangent 
of  half  the  inclination  of  the  equator,  and  n  is  the  projection  of  the 
north  pole,  Art.  (214). 

The  Equinoctial  colure  makes  with  the  primitive  plane  an 
angle  =  90°  +  23  J°.  Through  E'  draw  E'x  perpendicular  to  E'o. 
The  angle  GE'x  —  90°  +  231°,  and  O  =  tan  CEV  is  its  tan- 
gent and  EV  its  secant.  With  r  as  a  centre,  and  EV  as  a  radius 
describe  E'wE.  It  is  the  projection  of  the  half  of  the  equinoctial 
colure  above  the  primitive  plane.     It  must  pass  through  n. 

The  Solstitial  Colure,  being  perpendicular  to  EE',  passes 
through  the  point  of  sight  and  is  projected  into  the  right  line  pq. 

To  project  any  other  meridian,  as  that  which  makes  an  angle  of 
30°  with  the  equinoctial  colure  ;  produce  the  arc  E'nE  until  it 
intersects  O  produced.  The  point  of  intersection,  which  we  de- 
note by  *,  will  be  the  projection  of  the  south  pole,  and  since  all 
the  meridians  pass  through  the  poles,  their  projections  will  pass 
through  n  and  s,  and  ns  will  be  a  chord  common  to  the  projec- 
tions of  all  the  meridians.  If  at  its  middle  point  r,  the  perpen- 
dicular rl  be  erected,  this  will  contain  the  centres  of  all  these  pro- 
jections. If  through  n,  nl  be  drawn  making  ml  —  30°,  it  will  be 
the  radius  of  the  projection  of  that  meridian  which  makes  an  angle 
of  30°  with  the  equinoctial  colure,  since  rn  is  the  radius  of  the 
projection  of  this  colure  Art.  (211)  ;  and  I  is  the  centre  of  the 
projection  of  the  required  meridian,  and  ynz  the  projection. 

To  project  a  parallel  of  latitude,  as  the  Arctic  Circle  ;  lay  off 
Ei  =  47°,  the  sum  of  the  inclination,  23^°,  and  the  polar  distance, 
231  °  Art.  (207).  Draw  E'i ;  Co  =  tan  IE/,  and  o  is  one  ex- 
tremitv  of  a  diameter  of  the  projection.  Since  the  inclination  is 
equal  to  the  polar  distance,  the  other  extremity  is  at  C,  Art.  (207), 
and  the  circle  on  Co,  as  a  diameter,  is  the  required  projection. 


SPHKBICAL    PROJECTIONS.  129 

For  the  Tropic  of  Cancer  ;  lay  off  Ep  =  23^°  -f  66£°  ;  Cp  is 
the  tangent  of  half  Ep,  and  p  one  extremity  of  a  diameter  of  the 
projection.  From  k  lay  off  kh  =  66^°  the  polar  distance  of  the 
parallel.  Then  Eh  22s  43°  equal  the  difference  between  the  polar 
distance  and  the  inclination,  and  Cv  is  the  tangent  of  its  half,  and 
v  the  other  extremity  of  the  diameter,  and  the  circle  on  pv1  the 
required  projection.  The  projection  of  this  tropic  is  tangent  to 
the  ecliptic  at  p,  Art.  (65). 


216.  Since  each  point  on  the  hemisphere,  below  the  primitive 
plane,  has  a  greater  polar  distance  than  90°  and  will  therefore  be 
projected  without  the  primitive  circle,  Art.  (205),  and  those  cir- 
cles near  the  point  of  sight  will  thus  be  projected  into  very  large 
circles,  we  make  a  more  natural  representation  of  this  hemisphere 
by  revolving  it  180°,  as  in  the  orthographic  projection,  about  a 
tangent  at  E,  the  point  of  sight  being  moved  to  the  pole  of  the 
primitive  circle  in  its  new  position.  The  hemisphere  is  then 
above  the  primitive  plane  and  is  projected  as  in  the  preceding 
article;  s  being  the  projection  of  the  south  pole;  EmnE"  of  the 
other  half  of  the  equator;  EsE"  of  the  equinoctial  colure,  &c. 


217.  If  the  projection  be  made  on  any  other  great  circle  than 
the  ecliptic,  as  on  that  making  with  the  equator  an  angle  denoted 
by  A,  the  construction  will  be  the  same,  the  angle  A  being  used 
instead  of  23j°. 

If  on  the  horizon  of  a  place,  A  must  equal  90°  minus  the  lati- 
tude.    Art.  (193). 


218.  If  the  projection  be  made  on  the  equator,  the  preceding 
problem  is  much  simplified.  Thus  let  EpE'q,  Fig.  96,  be  the 
equator,  E  and.  E'  the  equinoctial  points.  EE'  is  the  intersection 
of  the  plane  of  the  ecliptic  with  that  of  the  equator  and  pq  is  its 


130  DESCRIPTIVE   GEOMETRY. 

line  of  measures  and  EoE'  its  projection,  m  being  the  centre  and 
mn  =  tan  231°. 

Since  the  meridians  pass  through  the  point  of  sight  they  are 
projected  into  right  lines.  EE'  is  the  projection  of  the  equinoc- 
tial and  pq  of  the  solstitial  colure  ;  and  ynz  of  the  meridian  which 
makes  an  angle  of  30°  with  the  solstitial  colure,  n  being  the  pro- 
jection of  the  north  pole* 

The  parallels  of  latitude,  being  parallel  to  the  equator,  are  pro- 
jected as  in  Art.  (208);  the  Arctic  Circle  into  arb;  and  the 
Tropic  of  Cancer  into  dof;  na  being  equal  to  tan  J(23-±-°) ;  and 
rid  =  tan  i(66J°). 


219.  If  the  projection  be  on  the  solstitial  colure  ;  let  ENE'S,  Fig. 
97,  be  the  primitive  circle;  EE'  its  intersection  with  the  plane  of 
the  equator. 

The  Equator,  being  perpendicular  to  the  primitive  plane,  passes 
through  the  point  of  sight  and  is  projected  into  EE'. 

The  Ecliptic,  for  the  same  reason,  is  projected  into  oo',  oCE  be- 
ing equal  to  231°. 

N^/S  is  the  projection  of  the  meridian,  making  with  the  primitive 
plane  an  angle  of  30°,  m  being  its  centre  and  Cm  —  tan  30°. 

adb  and  a'hb'  are  the  projections  of  the  polar  circles,  Cx  and 
Cx'  being  each  equal  to  the  secant  of  23j°,  and  xb  and  x'b'  each 
equal  to  the  tangent  of  23^-°,  Art.  (213). 

ogf  and  o'g'd'  are  the  projections  of  the  tropics,  each  described 
with  a  radius  equal  to  the  tan  661°. 


GLOBULAR    PROJECTIONS. 

220.  By  an  examination  of  an  orthographic  or  stereographic 
projection,  it  will  be  observed  that  the  projections  of  equal  arcs,  of 
great  circles  which  pass  through  the  pole  of  the  primitive  circle, 
are  very  unequal  in  length.     In  the  orthographic,  as  the  arc  is  re- 


SPHERICAL   PROJECTIONS.  131 

moved  from  the  pole  its  projection  is  diminished,  and  when  near 
the  primitive  circle  becomes  very  small,  while  the  reverse  is  the 
case  in  the  stereographic. 

To  avoid  this  inequality,  as  far  as  possible,  the  point  of  sight  is 
taken  in  the  axis  of  the  primitive  circle,  without  the  surface,  and  at 
a  distance  from  it  equal  to  the  sine  of  45°  =  R  y/\. 

Spherical  projections,  with  this  position  of  the  point  of  sight  are 
called  Globular. 

Thus  let  the  quadrant  E/>,  Fig.  98,  be  bisected  at  M,  and  S  the 
point  of  sight.  M  is  projected  at  m,  and  Cm  will  be  equal  to  mE, 
For  we  have  the  proportion  : 

oS  :  oM  : :  CS  :  Cm ;      whence,     Cm  =  — ^M  -  - .  (1). 

GO 

Since 

oM  a  oC  =  qS  ==  R -/IT 
we  have 

oS  =  R  +  2RV/|7         and  CS  =  R  +  R\/1 

Substituting  these  in  (1),  and  reducing,  we  have 

T> 

Cm  =  —  i=  mE  ; 

and  it  will  be  found  that  the  projections  of  any  other  t/o  CTdf«l 
arcs  of  this  quadrant  are  very  nearly  equal.  This  is  the  c<»V  ad- 
vantage of  this  mode  of  projection,  as  .the  projections  or  U.e  cir- 
cles of  the  sphere,  being  the  intersections  of  their  projecting  cone* 
with  the  primitive  plane  will,  in  general,  be  ellipses. 


GNOMONIC    PROJECTION. 

221.  If  the  sphere  be  projected  on  a  tangent  plane  at  any  point, 


132  DESCRIPTIVE    GKOMETRY. 

the  point  of  sight  being  at  its  centre*,  the  projection  is  called 
Gnomon  ic. 

In  this  case  the  projections  of  all  meridians  are  right  lines, 
since  their  planes  pass  through  the  point  of  sight. 

If  the  point  of  contact  be  on  the  equator  the  projections  of  the 
parallels  of  latitude  will  be  arcs  of  hyperbolas,  Art.  (165). 

If  the  point  of  contact  be  at  either  pole  of  the  earth,  these  pro- 
jections will  be  circles. 

By  this  mode  of  projection,  the  portions  of  the  sphere  distant 
from  the  point  of  contact  will  be  very  much  exaggerated. 


CYLINDRICAL    PROJECTION. 

222.  If  a  cylinder  be  passed  tangent  to  a  sphere  along  the 
equator,  and  the  point  of  sight  be  taken  at  the  centre  of  the 
sphere,  and  the  circles  of  the  sphere  be  projected  on  the  cylinder, 
and  the  cylinder  be  then  developed,  we  have  a  developed  projec- 
tion called  the  cylindrical  projection. 

In  this  case  the  meridians  will  be  projected  into  right  lines,  ele- 
ments of  the  cylinder,  which  are  developed  into  parallel  lines  per- 
pendicular to  the  developed  equator,  Art.  (161);  and  the  paral- 
lels into  circles  which  are  developed  into  right  lines  perpendicular 
to  the  developed  meridians,  and  at  distances  from  the  equator 
each  equal  to  the  tangent  of  the  latitude  of  the  parallel. 


CONIC    PROJECTION. 

223.  If  a  cone  be  passed  tangent  to  a  sphere  along  one  of  its 
parallels  of  latitude,  and  the  circles  of  the  sphere  be  projected  on 
it,  the  point  of  sight  being  at  the  centre,  and  the  cone  be  then 
developed,  we  have  a  developed  projection  called  the  conic  projec 
tion. 

In  this  case  the  meridians  will  be  projected  into  right  lines, 


SPHERICAL    PROJECTIONS.  133" 

elements  of  the  cone,  which  are  developed  into  right  lines  passing 
through  the  vertex ;  and  the  parallels  into  circles  whose  develop- 
ments will  be  arcs  described  from  the  vertex,  each  with  a  radius 
equal  to  the  distance  of  the  projection  from  the  vertex,  Art.  (166.) 

Thus,  in  Fig.  90,  let  EPE'  be  a  section  of  the  sphere  and  V  the 
vertex  of  the  cone  tangent  along  the  parallel  of  which  ab  is  the 
orthographic  projection.  The  equator  will  be  projected  into  a 
circle  whose  diameter  is  ee'  and  the  parallel  MJST  into  one  whose 
diameter  is  mn,  the  first  being  developed  into  an  arc  of  which  Ye 
is  the  radius,  and  the  second  into  one  of  which  Ym  is  the  radius. 
The  radius  Ya  of  the  development  of  the  circle  of  contact  is  evi- 
dently the  tangent  of  its  polar  distance. 

The  drawing  in  this,  as  in  the  cylindrical  projection,  for  those 
portions  of  the  sphere  distant  from  the  circle  of  contact,  will  evi- 
dently greatly  exaggerate  the  parts  projected. 


224.  This  exaggeration  may  be  lessened  by  making  the  projec- 
tion on  a  cone  passing  through  two  circumferences  equally  dis- 
tant, one  from  the  equator  and  the  other  from  the  pole.  Thus,  let 
ab  and  cd,  Fig.  100,  represent  two  circles  equally  distant  from  EE' 
and  P,  Ea  being  one  fourth  of  the  quadrant;  in  this  case  while  the 
parts  Etf  and  cP  will  be  exaggerated  in  projection,  the  part  ac  will 
be  lessened. 


225.  When  a  small  portion  of  the  surface,  between  two  given 
parallels,  is  to  be  represented,  the  conic  method  may  be  well  used, 
taking  the  cone  tangent  to  the  sphere  along  a  parallel  midway 
between  the  given  parallels,  if  the  first  method  be  used ;  or  pass- 
ing through  two  parallels,  each  distant  from  a  limiting  parallel 
one  fourth  the  arc  of  the  given  portion,  if  the  second  method  be 
used. 


134  DESCRIPTIVE    GEOMETRY. 


CONSTRUCTION    OF    MAPS.  , 

226.  If  it  be  desired  to  represent  the  entire  surface  of  the  earth 
by  a  map,  either  of  the  preceding  methods  may  be  used.  In  this 
case  it  is  usual  to  divide  the  quadrant,  from  the  pole  to  the  equa- 
tor, into  nine  equal  parts  and  project  the  parallels  of  latitude 
through  each  of  these  points,  as  well  as  the  polar  circles  and 
tropics;  and  also  to  divide  the  semi-equator  into  twelve  equal 
parts  and  to  project  the  meridians  passing  through  these  points. 
These  meridians  will  be  15°  apart  and  are  called  hour-circles. 

The  projections  of  the  different  points  to  be  represented  are 
then  made  and  the  map  filled  up  in  detail. 


227.  The  Stereographic  projection  gives  the  most  natural  rep- 
resentation and,  in  general,  is  of  the  easiest  construction. 

In  the  Globular,  when  the  equator  is  taken  as  the  primitive 
circle,  the  projections  of  the  meridians  are  right  lines,  and  of  the 
parallels  of  latitude,  circles;  and  this  projection  has  the  advantage 
that  these  parallels,  which  are  equally  distant  in  space,  have  their 
projections  also  very  nearly  equally  distant. 


228.  A  very  simple  construction,  when  the  primitive  circle  is  a 
meridian,  is  sometimes  made  thus :  Divide  the  arcs  EN,  E'N,  ES, 
and  E'S,  Fig.  101,  each  into  nine  equal  parts,  and  the  radii  CN" 
and  CS  also  each  into  nine  equal  parts,  then  describe  ares  of  cir- 
cles through  each  of  the  three  corresponding  points  of  division, 
for  the  representatives  of  the  parallels. 

In  the  same  way  divide  CE  and  CE',  each  into  six  equal  parts, 
and,  through  the  points  of  division  and  the  poles  N  and  S,  describe 
arcs  of  circles  for  the  representatives  of  the  meridians. 

This  representation,  though  called  the  equidistant  projection,  is 


SPHERICAL   PROJECTIONS.  135 

not   strictly   a   projection.     It   differs   little,  however,  from   the 
globular  projection* 


LORGNA  S    MAP. 

229.  A  map  of  the  globe  is  sometimes  made  by  describing  a 
primitive  circle  with  a  radius  equal  to  R^/2,  R  being  the  radius 
of  the  sphere,  and  regarding  this  as  the  representative  of  the 
equator.  Through  its  centre  draw  right  lines  making  angles  with 
each  other  of  15°,  for  the  meridians. 

To  represent  the  parallels :  With  the  centre  of  the  primitive  cir- 
cle as  a  centre  and  with  a  radius  equal  to  \/2llh,  h  being  the 
altitude  of  the  zone,  included  between  the  pole  and  the  parallel, 
describe  a  circle  to  represent  each  parallel  in  succession. 

The  area  of  the  primitive  circle  is  evidently  equal  to  the  area  oi 
the  hemisphere,  and  the  area  of  each  other  circle  to  that  of  the 
eorresponding  zone.  Hence,  the  area  between  any  two  circum- 
ferences will  be  equal  to  that  of  the  zone  included  between  the 
eorresponding  parallels. 

Also  the  area  of  any  quadrilateral  formed  by  the  arcs  of  two 
meridians  and  two  parallels  will  be  equal  to  its  representative  on 
the  primitive  plane. 


MERCATOR  S    CHART, 

230.  This  chart,  which  is  much  used  by  navigators,  is  a  modifi- 
cation of  the  cylindrical  projection.  It  has  the  great  advantage, 
that  the  course  of  a  ship  on  the  surface  of  the  sphere  which  makes 
a  constant  angle  with  the  meridians  intersected  by  it,  will  be  repre- 
sented on  the  chart  by  a  right  line, 

To  secure  this,  as  the  length  of  the  representative  of  a  degree 
of  longitude,  as  compared  with  the  arc  itself,  is  manifestly  in 
creased  as  the  distance  from  the  equator  increases,  it  is  necessar) 


•136  DESCRIPTIVE   GEOMETRY. 

that  the  representatives  of  the  degrees  of  latitude  should  increase 
in  the  same  ratio. 

But  the  length  of  a  degree  of  longitude,  at  any  latitude,  is 
known  to  be  equal  to  the  length  of  a  degree  at  the  equator  multi- 
plied by  the  cosine  of  the  latitude,  and  since  the  representative  of 
this  degree  at  all  latitudes  is  constant  on  the  chart,  being  the  dis 
tance  between  two  parallel  lines  the  representatives  of  two  consec- 
utive meridians,  it  follows  that  as  we  depart  from  the  equator,  this 
representative,  as  compared  with  the  arc  itself,  increases  as  the 
cosine  of  the  latitude  decreases,  or  increases  as  the  secant  in- 
creases, and  hence  the  representative  of  the  degree  of  latitude 
must  increase  in  the  same  ratio ;  that  is,  this  representative,  at 
any  given  latitude,  must  equal  its  length  at  the  equator  multiplied 
hy  the  natural  secant  of  this  latitude. 

By  adding  the  representatives  of  the  several  degrees,  or,  still 
more  accurately,  of  the  several  minutes  of  a  quadrant,  the  distance 
of  the  representative  of  each  parallel  from  the  equator  may  be 
found,  and  the  chart  may  then  be  thus  constructed.  Draw  a 
right  line  to  represent  the  equator,  then  a  system  of  equidistant  ' 
parallel  lines  for  the  meridians;  on  either  one  of  these  lay  off  the 
proper  distances  computed  as  above  for  each  paraliel  to  be  repre- 
sented, and  through  the  extremities  of  these  distances  draw  right 
lines  perpendicular  to  the  system  first  drawn.  They  represent  the 
parallels. 


231.  All  the  maps  constructed  as  in  the  preceding  articles, 
though  giving  a  general  representation  of  the  relative'  position  of 
objects  on  the  earth's  surface,  are  defective  in  this,  that  there  is  no 
definite  relation  between  actual  distances  of  points  and  the  repre- 
sentation of  these  distances  on  the  maps,  so  that  there  can  be  no 
scale  on  the  map  by  which  these  actual  distances  can  be  deter- 
mined. 

As  in  detailed  representations  of  smaller  portions  of  the  surface, 
this  scale  is  absolutely   necessary,   other  modes  of  constructing 


SPHERICAL    PROJECTIONS.  137 

these  maps  have  been  devised,  by  which  a  near  approximation  to 
an  accurate  scale  is  made. 


FLAMSTEAD  S    METHOD. 

232.  In  this  method,  modified  and  improved,  and  now  in  very 
general  use,  a  right  line,  AB,  Fig.  102,  is  drawn,  which  represents 
the  rectified  arc  of  the  meridian  passing  through  the  middle  of  the 
portion  to  be  mapped.  A  point,  c,  is  then  assumed  to  represent 
the  point  in  which  the  parallel  midway  between  the  extreme  par- 
allels intersects  this  meridian,  and  from  this  point,  in  both  direc- 
tions, equal  distances,  ex,  cy,  <fcc,  are  laid  off,  each  representing,  the 
true  length  of  one  degree  of  the  meridian.  Then  with  a  radius 
equal  to  the  tangent  of  the  polar  distance  of  this  central  parallel, 
the  arc  dee  is  described  to  represent  the  parallel.  This  arc  is  the 
development  of  the  line  of  contact  of  a  cone  tangent  to  the  sphere. 
Also,  with  the  same  centre,  arcs  are  described  through  each  of  the 
points  of  division  #,  y  ---r,  o,  to  represent  the  other  parallels  one 
degree  distant  from  each  other.  Then,  on  each  of  these  arcs, 
from  AB,  lav  off  both  ways  the  arcs  co,  ca,  y\\  yv',  os,  os\  &c, 
each  equal  to  the  length  of  a  degree  of  longitude  at  the  points 
c,  o,  <fcc,  viz,  the  length  of  a  degree  at  the  equator  multiplied  by 
the  natural  cosine  of  the  latitude  of  the  point.  Through  the 
points  a,  v,  s,  <fcc,  and  a\  v',  s\  <fcc,  draw  the  lines  avs,  and  a'v's'. 
They  will  represent  the  two  meridians  making  an  angle  of  one 
degree  each,  with  the  central  meridian. 

In  the  same  way,  the  representatives  of  the  two  meridians  next 
to  these  may  be  constructed,  by  laying  off  on  the  same  arcs  from 
a,  v,  s,  <fcc,  distances  each  equal  to  ca,  yv,  os,  <fcc,  and  drawiug 
lines  through  the  points  thus  determined,  and  so  on,  until  the 
representatives  of  the  extreme  meridians  of  the  portion  to  be 
mapped  are  drawn. 

In  this  map  the  scale  along  the  central  meridian  and  parallels 
will  be  accurate.     In  other  directions  when  the  map  does  not  rep- 


138  DESCRIPTIVE    GEOMETRY. 

resent  a  very  extended  portion  of  the  sphere,  a  verx  near  approxi- 
mation to  accuracy  is  made. 


THE    POLYCONIC    METHOD. 

233.  In  this  method,  by  which  the  elegant  an/  accurate  maps 
of  the  TJ.  S.  Coast  Survey  are  constructed,  the  central  meridian 
and  parallel  are  represented  as  in  the  precer'Xg  ari>!e  The 
representatives  of  the  other  parallels  are  each  acs^rioed,  through 
the  proper  point  of  division,  with  a  radius  equal  to  the  tangent  of 
its  polar  distance,  thus  giving  the  development  of  the  lines  of  con- 
tact of  so  many  tangent  cones  to  the  sphere.  The  length  of  each 
degree  of  longitude  is  then  laid  off  as  in  the  former  method  ami 
the  representatives  of  the  meridians  drawn. 

In  this  method,  also,  the  scale,  along  the  central  meridian  and 
parallels,  is  accurate  and  in  other  directions  very  nearly  so.  This 
has  the  advantage  that  the  representatives  of  the  meridians  and 
parallels  are  perpendicular  to  each  other  as  in  space,  which  is  not 
the  case  in  Flamstead's  method. 


234.  When  very  small  areas  are  mapped  this  method  is  thus 
modified  in  the  coast  survey  office.  The  same  process  is  used  as 
above  to  construct  the  representatives  of  the  meridians  with  accu- 
racy. Then  commencing  with  the  central  parallel,  the  distance 
ex,  Fig.  102,  between  it  and  the  consecutive  one,  as  measured  on 
AB,  is  laid  off  on  each  meridian  in  both  directions  and  through 
the  extremities  lines  drawn  to  represent  the  consecutive  parallels. 
Then  from  these  the  same  distances  are  laid  off  for  the  next  and  ■ 
so  on  until  all  are  constructed. 

The  first  set  of  parallels,  described  as  in  the  preceding  article, 
which  should  be  in  pencil,  are  then  erased. 

By  this  method  equal  meridian  distances  are  everywhere  in- 
cluded between  the  parallels,  and  the  scale  accurate  only  in  the 


8P1IKKICAL    PROJECTIONS.  139 

direction  of  the  meridians,  and  central  parallel.     This  is  called  the 
equidistant  poly  conic  method. 


235.  When  the  polar  distance  of  the  parallel  is  much  greater 
than  45°,  the  practical  construction  of  its  representative  becomes  * 
difficult,  as  its  centre  will  be  so  far  distant.  In  6uch  case,  for  both 
the  polyconic  method  and  that  of  Flamstead,  tables  are  carefully 
computed  giving  the  rectilineal  co-ordinates  of  the  points  of  the 
representatives  of  the  parallels,  for  each  minute  of  latitude  and 
longitude,  and  these  representatives  can  then  be  accurately  con- 
structed by  points.  The  tables  thus  computed  in  the  U.  S,  Coast 
Survey  office  are  very  much  extended  and  of  great  value. 


PART  III. 

SHADES  AND  SHADOWS. 


PRELIMINARY    DEFINITIONS. 

236.  To  represent  a  body  with  accuracy,  it  is  not  only  neces- 
sary that  the  drawing  should  give  the  representations  of  the  de- 
tails of  its  form,  but  also  of  its  colors,  whether  natural  or  artificial, 
or  the  effect  of  light  and  shade. 

Different  portions  of  the  same  body  will  appear  lighter  or  darker 
according  as  the  light  falls  directly  upon  it  or  is  excluded  from  it, 
by  itself  or  some  other  body.  A  simple  application  of  the 
elementary  principles  previously  deduced  will  enable  us  to  limit 
and  represent  these  portions,  and  constitutes  that  part  of  the 
subject  called  Shades  and  Shadows. 


237.  It  is  a  principle  of  Optics,  that  the  effect  of  light,  when 
in  the  same  medium  and  unobstructed,  is  transmitted  from  each 
point  of  a  luminous  body  in  every  direction,  along  right  lines. 
These  right  lines  are  called  rays  of  light.  Any  right  line,  there- 
fore, drawn  from  a  point  of  a  luminous  body  will  be  regarded  as  a 
ray  of  light. 

The  sun  is  the  luminous  body  which  is  the  principal  source  of 
light.  It  is  at  so  great  a  distance,  that  rays  drawn  from  any  of 
its  points  to  an  object  on  the  earth's  surface,  may,  without  mate- 


STIADKS    AND    SHADOWS.  141 

rial  error,  be  regarded  as  parallel.     In  the  construction  of  prob- 
lems, in  this  part  of  our  subject,  they  will  be  so  taken. 


238.  Let  SR,  Fig.  103,  indicate  the  direction  of  the  parallel 
rays  of  light,  coming  from  the  source  and  falling  on  the  opaque 
body  B ;  and  let  n-mlop-q  be  a  cylinder,  whose  rectilinear  elements 
are  rays  of  light  parallel  to  SR,  enveloping  and  touching  this  body. 

It  is  evident  that  all  light,  coming  directly  from  the  source,  will 
be  excluded  from  that  part  of  the  interior  of  this  cylinder  which  is 
behind  B.  This  part  of  space  from  which  the  light  is  thus  ex- 
cluded by  the  opaque  body,  is  the  indefinite  shadow  of  the  body; 
and  any  object  within  this  portion  of  the  cylinder  is  in  the  shadow, 
or  has  a  shadow  cast  upon  it. 

The  line  of  contact  mlop  separates  the  surface  of  the  opaque 
body  into  two  parts.  That  part  which  is  towards  the  source  01 
light,  and  on  which  the  rays  fall  directly,  is  the  illuminated  part ; 
and  that  part  opposite  the  source  of  light  from  which  the  rays  are 
excluded  by  the  body  itself,  is  the  shade  of  the  body. 

This  line  of  contact  of  the  tangent  cylinder  of  rays,  thus  sepa- 
rating the  illuminated  part  from  the  shade,  js  the  line  of  shade. 
Any  plane  tangent  to  this  cylinder  of  rays  will  also  be  tangent  to 
the  o'paque  body  at  some  point  of  this  line  of  contact,  Art.  (116); 
and,  conversely,  every  p^ne  tangent  to  the  opaque  body  at  a  point 
of  the  line  of  shade  will  be  tangent  to  this  cylinder,  and  contain  a 
rectilinear  element,  Art.  (110),  or  ray  of  light,  and  thus  be  a 
plane  of  rays.  Hence  points  of  the  line  of  shade  on  any  opaque 
body  may  be  obtained  by  passing  planes  of  rays  tangent  to  the  body, 
and  finding  their  points  of  contact. 


239.  If  the  cylinder  n-mlop-q  be  intersected  by  any  surface, 
as  the  plane  ABCD,  that  part  m'l'o'p'  of  this  plane,  which  is  in 
the  shadow,  is  the  shadow  of  B  on  this  plane.  That  is,  the  shadow 
of  an  opaque  body  on  a  surface  is  that  part  of  the  surface  fron 


142  DESCRIPTIVE   GEOMETRY. 

which  the  rays  of  light  are  excluded  by  the  interposition  of  this 
body  between  it  and  the  source  of  light. 

The  bounding  line  of  this  shadow,  or  the  line  which  separata 
the  shadow  on  a  surface  from  the  illuminated  part  of  that  surface, 
is  the  line  of  shadow.  It  is  also  the  line  of  intersection  of  the  cyl- 
inder of  rays  which  envelops  the  opaque  body,  and  the  surface 
on  which  the  shadow  is  cast. 


240.  When  the  opaque  body  is  bounded  by  planes,  the  cylinder 
of  rays,  by  which  its  shadow  is  determined,  will  be  changed  into 
several  planes  of  rays,  which  will  include  the  indefinite  shadow ; 
and  the  line  of  shade  will  be  made  up  of  right  lines  which,  though 
•  not  lines  of  mathematical  tangency,  are  the  outer  lines  in  which 
these  planes  touch  the  opaque  body,  and  still  separate  the  illu- 
minated part  from  the  shade. 

The  line  of  shadow,  in  this  case,  is  also  made  up  of  the  lines  of 
intersection  of  these  planes  with  the  surface  on  which  the  shadow 
is  cast,  and  still  separates  the  illuminated  part  of  this  surface  from 
the  shadow. 


SHADOW    OF    POINTS    AND    LINES. 

241.  The  indefinite  shadow  of  a  point  may  be  regarded  as  that 
part  of  a  ray  of  light,  drawn  through  it,  which  lies  in  the  direc- 
tion opposite  to  that  of  the  source  of  light,  and  the  point  in  which 
this  ray  pierces  any  surface  is  the  shadow  of  the  point  on  that 
surface. 


242.  The  shadow  of  a  right  line  will  then  be  determined  by 
drawing  through  its  different  points  rays  of  light.  These  form  a 
plane  of  rays,  and  the  intersection  of  this  plane  with  any  surface 
is  the  shadow  of  the  line  on  that  surface. 


SHADES    AND   SHADOWS.  14:3 

To  determine  whether  a  given  plane  is  a  plane  of  lays,  it  is  only 
necessary  to  ascertain  whether  it  contains  a  ray  of  light.  This 
may  be  done  by  drawing  through  any  one  of  its  points  a  ray.  If 
this  pierces  the  planes  of  projection  in  the  traces  of  the  given 
plane,  it  is  a  plane  of  rays,  Art.  (30). 


243.  The  shadow  of  a  right  line  on  a  plane  may  be  constructed 
by  finding  the  shadows  cast  by  any  two  of  its  points  on  the  plane, 
Art.  (241),  and  joining  these  by  a  right  line;  or  by  joining  the 
shadow  of  any  one  of  its  points  with  the  point  in  which  the  line 
pierces  the  plane,  Art.  (30). 

If  the  right  line  be  parallel  to  the  planey  its  shadow  on  the  plane 
will  be  parallel  to  the  line  itself,  and  the  shadow  of  a  definite  por- 
tion of  it  will  be  equal  to  this  portion,  Art.  (14). 

If  the  line  is  a  ray  of  light,  its  shadow  on  any  surface  will  be 
a  point. 

Thus  let  MN,  Fig.  104,  be  a  limited  right  line,  and  tTt'  the 
plane  on  which  its  shadow  is  required,  MR  indicating  the  direction 
of  the  rays  of  light. 

Through  the  extremities  M  and  N,  draw  the  rays  MR  and  NS, 
They  pierce  tTt'  in  R  and  S,  Art.  (40),  which  are  points  of  the 
shadow,  Art.  (241).  Join  these  points  by  RS,  it  will  be  the  re- 
quired shadow. 

If  the  shadow  be  cast  on  the  horizontal  plane,  the  rays  through 
the  extremities  M  and  N,  Fig.  (a),  pierce  H  at  m,"  and  n'\  and 
m"n"  is  the  shadow. 

If  the  shadow  be  cast  on  the  vertical  plane,  these  rays  pierce  V 
at  m"  and  w",  Fig.  (6),  and  m"n"  is  the  shadow. 

If  MN  is  parallel  to  either  plane,  then  RS  in  Fig.  104,  and 
m"n"  in  Figs,  (a)  and  (b)  will  be  equal  to  MN. 

If  right  lines  are  parallel,  their  shadows  on  a  plane  are  parallel, 
since  they  are  the  intersections  of  parallel  planes  of  rays  by  this 
plane. 


lM  DESCRIPTIVE    GEOMETRY. 

244.  The  shadow  of  a  curve  will  be  determined  by  passing 
through  it  a  cylinder  of  rays.  Tlie  intersection  of  this  cylinder 
with  any  surface  will  be  the  shadow  of  the  curve  on  that 
surface. 

The  line  of  shadow  on  a  surface,  Art.  (239),  will  always  be  the 
shadow  of  the  line  of  shade ;  and  if  through  any  point  of  the 
line  of  shadow  a  ray  be  drawn  to  the  source  of  light,  it  will 
intersect  the  line  of  shade  in  a  point  which  casts  the  assumed 
point  of  shadow. 

From  this  it  follows  that  the  point  of  shadow  cast  by  any  line  in 
Space  upon  any  other  line,  may  be  found  by  constructing  the 
shadows  of  both  lines  on  a  plane ;  and  drawing  through  the  point 
of  intersection  of  these  shadows  a  ray,  it  will  intersect  the  second 
line  in  the  point  of  shadow  cast  upon  it,  by  the  point  in  which  it 
intersects  the  first. 


245.  If  two  lines  are  tangent  in  space,  their  shadows,  on  any 
surface,  will  be  tangent  at  the  point  of  shadow  cast  by  the  given 
point  of  contact.  For  the  cylinders  of  rays  through  the  lines  will 
be  tangent  along  the  ray  passing  through  the  point  of  contact, 
Art.  (117);  hence  their  intersections  by  any  surface  will  be  tan- 
gent at  the  point  where  this  ray  pierces  the  surface.  These  inter- 
sections are  the  shadows  of  the  lines. 


246.  The  shadow  of  a  curve  of  single  curvature,  on  a  plane  to 
which  it  is  parallel,  will  be  an  equal  curve,  since  each  of  its 
elements  will  cast  a  parallel  and  equal  element  of  the  shadow, 
Art.  (243). 

If  the  plane  of  the  curve  be  a  plane  of  rays,  its  shadow  on  a 
plane  will  evidently  be  a  right  line. 

The  shadow  of  the  circumference  of  a  circle,  on  a  plane  to  which 
it  is  parallel,  will  be  an  equal  circumference,  whose  centre  is  tb 
shadow  cast  by  the  given  centre. 


SHADKS   AND   SHADOW?.  145 

Also,  when  the  plane  on  which  the  shadow  is  cast  makes  a  sub' 
contrary  section  in  the  cylinder  of  rays  through  the  circumference, 
Art.  (206). 

In  all  other  positions,  when  its  plane  is  not  a  plane  of  raysr  its 
shadow  will  be  an  ellipse,  Art.  (160),  and  any  two  diameters  of 
the  circle  perpendicular  to  each  other  will  cast  conjugate  diameters 
of  the  ellipse  of  shadow.  For  if  at  the  extremities  of  either 
diameter  tangents  be  drawn,  they  will  be  parallel  to  the  other 
diameter,  and  their  shadow  parallel  to  the  shadow  of  this 
diameter,  Art.  (243).  But  the  shadows  of  the  two  tangents  are 
tangent  to  the  shadows  of  the  circle,  at  the  extremities  of  the 
shadow  of  the  first  diameter,  Art.  (245) ;  hence  the  shadow  of  the 
second  diameter  is  parallel  to  the  tangents  at  the  extremities  of 
the  shadow  of  the  first.  These  shadows  are  therefore  conjugate 
diameters  of  the  ellipse,  and  with  them  the  ellipse  may  be  con- 
structed.    Ana.  Geo.,  Art.  (150). 

If  the  shadows  of  two  perpendicular  diameters  be  found,  also 
perpendicular  to  each  other,  they  will  be  the  axes  of  the  ellipse, 
which  may  be  constructed  as  in  Art.  (59). 


BRILLIANT    POINTS. 

247.  In  looking  upon  the  illuminated  part  of  a  curved  surface, 
it  will  be  observed  that,  in  general,  one  or  more  points  appear 
much  more  brilliant  than  the  others.  This  is  due  to  the  fact  that 
the  ray  of  light,  incident  at  each  of  these  points,  is  reflected  imme- 
diately to  the  point  of  sight. 

These  points  are  called  brilliant  points  ;  and  since  it  is  a  prin- 
ciple of  Optics  that  the  incident  and  reflected  rays,  at  any  point  oi 
a  surface,  lie  in  the  same  normal  plane  on  opposite  sides  of  the 
normal  at  this  point,  and  making  equal  angles  with  it,  it  follow 
that  at  a  brilliant  point,  the  ray  of  light  and  the  right  line  draw 
to  the  point  of  sight  must  fulfil  these  conditions. 

When  the  point  of  sight  is  at  an  infinite  distance,  as  in  th 
10 


146  DESCRIPTIVE    GEOMETRY. 

Orthographic  projection,  the  brilliant  point  may  be  constructed 
thus :  Through  any  point  draw  a  ray  of  light  and  a  right  line  to 
the  point  of  sight.  These  will  be  parallel,  respectively,  to  the  cor- 
responding lines  at  the  required  point,  and  make  the  same  angle. 
Bisect  the  angle  formed  by  these  auxiliary  lines  as  in  Art.  (37). 
The  bisecting  line  will  be  parallel  to  the  line  which  bisects  the 
parallel  and  equal  angle  at  the  brilliant  point.  Hence,  perpendicu- 
lar to  this  bisecting  line,  and  tangent  to  the  surface^  pass  a  plane. 
Its  point  of  contact  will  be  the  brilliant  point ;  for  a  line  through 
this  point  parallel  to  the  bisecting  line  will  be  a  normal  to  the 
surface,  and  will  bisect  the  angle  formed  by  a  ray  of  light  and 
right  line  drawn  from  this  point  to  the  point  of  sight. 


PRACTICAL   PROBLEMS. 

248.  Problem  61.  To  construct  the  shadow  of  a  rectangular 
pillar,  on  the  planes  of  projection.  Let  mnfo,  Fig.  105,  be  the 
horizontal  and  m'n'p'q*  the  vertical  projection  of  the  pillar,  and 
let  (nn",  n'r)  indicate  the  direction  of  the  rays  of  light. 

The  upper  base  vertically  projected  in  m'n\  and  the  two  ver- 
tical faces  horizontally  projected  in  mn  and  mo,  are  evidently  illu- 
minated, and  together  form  the  illuminated  part  of  the  pillar;  and 
the  edges  (n,  pV),  (nl,  »'),  (lo,  n'm'),  and  (o,  q'm')  separate 
the  illuminated  part  from  the  shade,  and  make  up  the  line  of 
Bhade,  Art.  (240), 

Since  n  is  in  the  horizontal  plane  and  n"  the  shadow  of  N,  nn1' 
is  the  shadow  of  (n,  p'n')  on  H,  Art.  (243).  The  plane  of  rays 
through  (nl,  n')  is  perpendicular  to  V,  and  n"r,  parallel  to  nl,  is 
its  horizontal  and  rl"  its  vertical  trace ;  hence  n''r  is  that  part  of 
the  shadow  of  (n/,  n')  which,  is  on  H ;  and  rl",  limited  by  the  ray 
through  {In'),  that  part  on  V.  The  ray  (rx,  rn')  intersects  (nl,  n') 
in  (xnr),  and  nx  is  the  horizontal  projection  of  that  part  which, 
casts  the  equal  shadow  n"r)  Art.  (243),  and  xl  of  that  part  which 
casts  rV\ 


SHADES   AND    SHADOWS.  147 

l"o,r  is  the  shadow  of  (o/,  m'nf)  on  V,  parallel  and  equal  to 
itself. 

The  plane  of  rays  through  (o,  q'm')  is  perpendicular  to  H,  and 
os  is  its  horizontal  and  so"  its  vertical  trace ;  and  os  the  shadow  of 
(o,  q'm')  on  H,  and  so"  its  shadow  on  V,  q'y'  is  the  vertical  pro- 
jection of  the  part  which  casts  the  shadow  os  and  y'm'  of  that 
part  which  casts  its  equal  so",  Art.  (243). 

The  line  of  shadow  is  thus  the  broken  line  nn"r...o"...o1  and 
the  portion  of  the  planes  within  this  line  is  the  required  shadow, 
and  should  be  darkened  as  in  the  drawing. 

249.  Problem  62.  To  construct  the  shadow  of  a  rectangular 
abacus  on  the  faces  of  a  rectangular  pillar \ 

Let  mnlo,  Fig.  106,  be  the  horizontal,  and  m'n'Vo*  the  vertical 
projection  of  the  abacus,  and  cdd'c'  the  horizontal,  and  c'd'e'f  the 
vertical  projection  of  the  pillar;  MR  indicating  the  direction  of 
the  rays. 

The  lines  of  shade  on  the  abacus,  which  cast  the  required 
shadows,  are  evidently  the  two  edges  (mo,  m')  and  MN. 

The  plane  of  rays  through  (mo,  m')  is  perpendicular  to  V,  and 
intersects  the  side  face,  horizontally  projected  in  cc\  in  a  right  line 
perpendicular  to  V  at  s',  which  is  the  shadow  on  this  face,  m'r' 
is  the  vertical  trace  of  this  plane.  The  ray  MR,  through  M, 
pierces  the  front  face  of  ttfe  pillar  in  R,  Art.  (40),  the  shadow 
of  M,  and  (cr,  s'r')  is  the  shadow  of  the  part  (pm,  m')  on  this  face. 

MN  being  parallel  to  this  face,  its  shadow  on  it  will  be  parallel 
to  itself,  and  pass  through  R ;  hence  (rd,  r'x')  is  the  required 
shadow  cast  by  its  equal  MQ. 


250.  Problem  63.  To  Construct  the  shadow  of  an  upright  cross 
upon  the  horizontal  plane  and  upon  itself. 

Let  mnop,Y\g.  107,  be  the  horizontal,  and  c'n'r'g'f  the  vertical 
projection  of  the  cross. 

cc"  is  the  shadow  of  the  edge  (c,  c'd'),  c"  being  the  shadow  of 


143  DESCRIPTIVE    GEOMETRY. 

(cd')  ;  c"n"  is  the  shadow  of  its  equal  (en,  d'nf),  Art.  (243)  ;  n"h" 
of  the  edge  (n,  n'h')  ;  h"o"  of  its  equal  (no,  h')  ;  o"y"  of  (oy,  h'x')  ; 
(xy,  x')  of  a  part  of  (cd,  r'),  on  the  face  of  the  cross  vertically  pro- 
jected in  l'k'.  y"r"  is  the  shadow  of  the  remaining  part  of  (cd,  r) 
on  H.  (ex,  l'x')  is  the  shadow  of  (c,  l'r')  on  the  face  vertically  pro- 
jected in  l'h',  this  being  the  intersection  of  a  vertical  plane  of  rays 
through  (c,  I'?-')  with  this  face.  r"q"  is  the  shadow  of  its  equal 
(de,  r'q') ;  q"lc"  of  (e,  k'q')  ;  k"p"  of  (pe,  h' y')  ;  p"m"  of  (p,  m'g') ; 
and  m"t  of  a  part  of  (pm,  m). 

All  within  the  broken  line  thus  determined  on  II  is  darkened,  as 
also  the  part  cxyd,  the  horizontal  projection  of  the  shadow  cast  by 
the,  upper  part  of  the  cross  on  the  face  vertically  projected  in  l'h'. 


251.  Problem  64.  To  construct  the  shade  of  a  cylindrical 
column,  and  of  its  cylindrical  abacus,  and  the  shadow  of  the  abacus 
on  the  column. 

Let  mlo,  Fig.  108,  be  the  horizontal,  and  m'o'h'n'  the  vertical 
projection  of  the  abacus,  cde  the  horizontal,  and  cee'g'  the  vertical 
projection  of  the  column. 

Pass  two  planes  of  rays  tangent  to  the  column.  Since  each  con- 
tains a  rectilinear  element,  they  will  be  perpendicular  to  H  ;  and 
Id  and  kf,  parallel  to  the  horizontal  projection  of  the  ray  of  light, 
will  be  their  horizontal  traces,  Art.  (123);  and  (d,  z'd')  and  (/,  f'r') 
the  elements  of  contact,  which  are  indefinite  lines,  or  elements  of 
shade,  Art.  (238.) 

In  the  same  way  the  elements  of  shade  (i,  u'i'),  <fcc,  on  the 
abacus,  are  determined. 

The  line  of  shadow  on  the  column  will  be  cast  by  a  portion  of  the 
lower  circumference  of  the  abacus  towards  the  source  of  light. 
To  determine  any  point  of  it,  pass  a  vertical  plane  of  rays,  as  that 
whose  horizontal  trace  is  yx.  It  intersects  the  column  in  a  recti- , 
linear  element  (x,  x"x'),  and  the  circumference  in  a  point  Y 
Through  this  point  draw  the  ray  YX.  It  intersects  the  elemen  : 
in  X,  a  point  of  the  required  shadow,  Art.  (241)  •  and  in  the  same 


SHADES  AND    SHADOWS.  14:9 

way  any  number  of  points  may  be  found.  C  is  the  shadow  of  Q, 
and  D  of  L,  and  c'x'd'  is  the  vertical,  and  cxd  the  horizontal  pro- 
jection of  that  part  of  the  curve  of  shadow  which  can  be  seen. 

That  part  of  the  shade  on  the  abacus  and  column,  and  of  the 
shadow  on  the  column,  which  can  be  seen,  i3  darkened  in  the 
drawing. 


252.  Problem  65.  To  construct  the  shade  of  a  right  cylinder 
with  a  circular  base,  and  its  shadow  on  the  horizontal  plane  and  on 
its  interior  surface. 

Let  mlo,  Fig.  109,  be  the  base  of  the  cylinder,  and  (c,  d'c') 
its  axis. 

The  two  planes  of  rays,  whose  horizontal  traces  are  11"  and  kk", 
determine  the  two  elements  of  shade  {I,  n'V)  and  (&,  r'k'). 

These  elements  cast  the  shadows  11"  and  kk"  on  H,  Art.  (243). 

The  semi-circumference  of  the  upper  base  LUK  casts  its  shadow 
on  the  interior  of  the  cylinder,  and  the  other  semi-circumference 
on  the  horizontal  plane,  without  the  cylinder. 

The  shadow  of  the  latter  is  the  equal  semi-circumference  k"o"l" 
whose  centre  is  c" ,  Art.  (246),  and  this,  with  the  lines  11"  and  kk" , 
limits  the  shadow  of  the  cylinder  on  H. 

To  determine  the  shadow  on  the  interior  surface,  pass  any  vertical 
plane  of  rays,  as  that  whose  horizontal  trace  is  yx.  It  intersects 
the  cylinder  in  the  element  (x,  x"x')  and  the  semi-circumference 
in  the  point  Y.  The  ray  of  light  YX  intersects  (%,  x"x')  in  X,  a 
point  of  the  required  shadow  ;  and  in  the  same  way  any  number 
of  points  may  be  determined. 

The  shadow  upon  any  rectilinear  element,  may  be  found  by 
using  an  auxiliary  plane  which  shall  contain  this  element.  Thus 
Z  is  the  shadow  of  U  on  the  element  (z,  z "z'),  s"  of  S,  and  t" 
ofT. 

This  shadow  evidently  begins  at  L  and  K,  the  upper  ends  of 
the  elements  of  shade. 

The  direction  of  the  rays  is  so  taken  that  a  part  of  this  semi- 


150  DESCRIPTIVE    GEOMETRY. 

circumference  casts  its  shadow  on  H,  within  the  cylinder.  This 
part  is  horizontally  projected  in  st,  and  its  shadow  is  the  equal 
arc  s"t"  the  continuation  oik"o"l". 

Should  this  shadow  not  reach  H,  its  lowest  point  will  be  obtained 
by  using  as  an  auxiliary  plane  that  one  which  contains  the  axis ; 
and  the  point  in  which  the  vertical  projection  of  the  curve  of 
shadow  is  tangent  to  o'p',  by  using  the  plane  which  contains  the 
element  (o,  o'p'). 

That  part  of  the  shade  between  the  elements  (I,  n'V)  and  (o,  o'p') 
only,  can  be  seen  when  looking  on  V,  and  is  darkened  in  the 
drawing. 


253.  Problem  66.  To  construct  the  shade  of  an  oblique  cone 
and  its  shadow  on  the  horizontal  plane. 

Let  the  cone  be  given  as  in  Fig.  110,  S  being  its  vertex. 

If  two  planes  of  rays  be  passed  tangent  to  the  cone,  the  elements 
of  contact  will  be  the  elements  of  shade,  Art.  (238).  Since  these 
planes  must  pass  through  S,  Art.  (130),  they  must  contain  the 
ray  of  light  SR.  This  pierces  H  at  r,  and  rn  and  rp  are  the  hor- 
izontal traces  of  these  tangent  planes,  and  SN  and  SP  are  the  ele- 
ments of  contact,  Art.  (130). 

These  elements  cast  the  shadows  nr  and  pr,  which  limit  the 
shadow  of  the  cone  on  H. 

In  looking  upon  H,  the  shade  between  the  elements  SP  and  SQ 
only  is  seen ;  and  in  looking  upon  V,  that  between  SN  and  SO. 


254.  Problem  67.  To  construct  the  shade  of  an  ellipsoid  oj 
revolution  and  its  shadow  on  the  horizontal  plane. 

Let  the  ellipsoid  be  given  as  in  Fig.  111. 

The  line  of  shade  is  the  line  of  contact  of  a  cylinder  of  rays 
tangent  to  the  surface,  and  is  an  ellipse,  Ana.  Geo.,  Art.  (233) ; 
and  since  the  meridian  plane  of  rays,  whose  horizontal  trace  is 
cs,  is  evidently  a  principal  plane,  Art.  (206),  of  both  the  ellipsoid 


SHADES   AND   SHADOWS.  151 

and  cylinder,  bisecting  all  chords  of  both  surfaces  perpendicular 
to  it,  its  intersection  with  the  plane  of  the  ellipse  will  also  bisect 
all  chords  of  the  ellipse  perpendicular  to  it,  and  be  an  axis  of  the 
curve. 

This  meridian  plane  cuts  from  the  ellipsoid  a  meridian  curve, 
and  from  the  cylinder  two  rectilinear  elements,  rays  of  light,  tan- 
gent to  this  curve,  at  the  vertices  of  the  axis,  the  highest  and 
lowest  points  of  the  curve  of  shade. 

To  determine  these  points,  revolve  the  meridian  plane  'about 
(c,  eW),  until  it  becomes  parallel  to  V.  The  meridian  curve  will 
then  be  vertically  projected  into  e'm'd'n'^  CT  will  be  the  revolved 
position  of  CS,  the  axis  of  the  cylinder,  and  the  two  tangents  at 
r'  and  x\  parallel  to  cY,  will  be  the  vertical  projections,  in  revolved 
position,  of  the  two  elements  cut  from  the  cylinder,  and  R  and  X 
will  be  the  revolved  position  of  the  two  vertices.  When  the 
plane  is  revolved  to  its  true  position,  these  points  will  be  at 
P  and  Y,  Art.  (107),  and  PY  is  the  transverse  axis  of  the  required 
curve. 

The  conjugate  axis,  being  perpendicular  to  the  meridian  plane 
of  rays,  is  parallel  to  H,  and  therefore  horizontally  projected  into 
fg,  perpendicular  to  yp,  Art.  (36),  and  vertically  into  fg\ 

The  projections  of  this  curve  are  ellipses.  Of  the  horizontal  pro- 
jection, fg  is  evidently  the  longest  diameter,  and  therefore  the 
transverse  axis,  and  yp  the  conjugate,  and  the  ellipse  fpgy  is  the 
horizontal  projection  of  the  curve  of  shade. 

Since  the  tangents  to  the  curve  at  the  highest  and  lowest  points 
P  and  Y  are  horizontal,  their  vertical  projections  must  be  rTp'  and 
x'y'  tangent  to  the  vertical  projection  of  the  curve  at  p'  and  y\ 
These  tangents  being  parallel  tofg\p'y1  and  fg'  must  be  conju- 
gate diameters  of  the  ellipse  which  is  the  vertical  projection  of 
the  curve ;  hence  pyf'y'g\  on  these  diameters,  is  the  required  ver- 
tical projection. 

Points  of  this  curve  may  also  be  found  otherwise,  by  intersect 
ing  the  surface  by  any  horizontal  plane,  as  that  of  which  l'k'  i 
the  vertical  trace.     It  cuts   from  the   surface  a  circumference, 


152  DESCRIPTIVE   GEOMETRY. 

and  from  the  plane  of  the  curve  a  line  parallel  to  FG.  These 
lines  intersect  in  the  required  points.  K  and  L  are  thus  deter- 
mined. 

The  points  of  tangency,  ur  and  w\  are  the  vertical  projections 
of  the  points  in  which  the  curve  of  shade  is  intersected  by  the 
meridian  plane  parallel  to  V. 

The  shadow  cast  by  the  curve  of  shade  on  H  is  an  ellipse,  the 
intersection  of  the  tangent  cylinder  with  H.  The  conjugate  axis 
of  this  ellipse,/"^",  is  the  shadow  of  FG  parallel  and  equal  to 
itself,  and  the  semi-transverse  axis,  y"s,  the  shadow  of  CY,  since 
these  shadows  are  perpendicular  to  each  other,  and  y"s  bisects  all 
chords  perpendicular  to  it. 


255.  If  the  ellipsoid  becomes  a  sphere,  the  construction  of  the 
curve  of  shade  is  simplified,  as  it  then  becomes  the  circumference 
of  a  great  circle  perpendicular  to  CST  Art.  (121),  and  fg  is  the 
projection  of  that  diameter  which  is  parallel  to  H,  and  py  the 
projection  of  the  one  which  is  perpendicular  to  it,  Art.  (197),  and 
the  vertical  projection  may  be  obtained  in  the  same  way. 


256.  If  a  plane  of  rays  be  passed  tangent  to  any  surface  of 
revolution,  the  point  of  contact  will  be  a  point  of  the  curve  of 
shade,  Art.  (238).  If  through  this  point  of  contact  a  meridian 
plane  be  passed,  it  will  be  perpendicular  to  the  tangent  plane, 
Art.  (115),  and  cut  from  it  a  line  tangent  to  the  meridian  curve  at 
this  point,  Art.  (108).  The  plane,  which  projects  a  ray  of  light 
on  to  this  meridian  plane,  is  also  a  plane  of  rays  perpendicular  to 
it,  and  therefore  parallel  to  the  tangent  plane,  and  its  intersection 
with  the  meridian  plane  will  be  parallel  to  the  tangent  cut  from 
the  tangent  plane.  But  this  intersection  is  the  projection  of  the 
ray.  Hence,  if  a  tangent  be  drawn  to  any  meridian  curve  of  a 
surface  of  revolution  'parallel  to  the  projection  of  a  ray  of  light  on 
the  meridian  plane,  its  point  of  contact  will  be  a  point  of  the  line  oj 


SHADES    AND   SHADOWS.  153 

shade,  and  thus  any  number  of  points  may  be  found  on  different 
meridians,  and  the  line  of  shade  drawn. 

The  points  U  and  W,  in  Fig.  Ill,  may  thus  be  determined  by 
drawing  tangents  to  the  meridian  curve  which  is  parallel  to  V, 
para,  lei  to  cV. 


257.  Problem  68.  To  find  the  brilliant  point  on  any  surface 
of  revolution. 

Let  the  surface  be  an  ellipsoid,  given  as  in  Fig.  111.  Through  any 
point  of  the  axis  as  C,  draw  the  ray  of  light  CS  and  the  line  (co,  c') 
to  the  point  of  sight  in  front  of  the  vertical  plane.  Bisect  their 
included  angle  as  in  Art.  (37)  by  the  line  CQ.  To  determine  a 
plane  perpendicular  to  this  line  and  tangent  to  the  surface,  Art. 
(247),  through  it  pass  the  meridian  plane  of  which  cq  is  the 
horizontal  trace.  This  plane  cuts  from  the  surface  a  meridian 
curve,  and  from  the  required  tangent  plane  a  right  line  tangent  to 
this  curve  at  the  required  point,  and  perpendicular  to  CQ. 

Revolve  this  plane  about  (c,  cd')  until  it  becomes  parallel  to  V. 
The  meridian  curve  will  be  vertically  projected  in  e'n'd'm',  and 
CQ  in  c'q"\  Art.  (29).  Tangent  to  e'm'd'  and  perpendicular  to 
cq"'  draw  i'z'"',  z'"  is  the  vertical  projection  of  the  revolved 
position  of  the  required  point,  horizontally  projected  at  z",  and 
when  the  plane  is  revolved  to  its  true  position  at  z.  Z  is  then  the 
required  brilliant  point. 


258.  Problem  69.  To  construct  the  shade  and  shadow  of  an 
upright  screw. 

Let  ecf,  Fig.  112,  be  the  circular  base  of  a  right  cylinder  whose 
axis  is  oc\  and  let  em'e'  be  an  isosceles  triangle  in  the  plane  of  the 
element  ee  and  of  the  axis,  its  base  coincident  with  the  element. 
If  this  triangle  be  moved  about  the  cylinder,  its  plane  always 
containing  the  axis,  with  a  uniform  angular  motion,  and  at  the 
game  time  with  a  uniform  motion  in  the  direction  of  the  axis,  so 


154  PESCRIFflVK   GEOMETRY. 

that  after  passing  around  once  it  will  occupy  the  position  efm"e\ 
it  will  generate  a  volume  called  the  thread  of  a  screw. 

The  cylinder  is  the  cylinder  of  the  screw. 

It  is  evident  that  the  two  sides  m'e  and  m'e'  will  each  generate 
a  portion  of  a  helicoid,  Art.  (92),  the  side  m'e'  generating  the 
tipper  surface  of  the  thread,  and  m'e  the  lower  surface. 

The  point  rn'i  generates  the  outer  helix  of  the  thread,  and  the 
point  e  the  inner  helix. 

The  curve  of  shade  on  the  lower  surface  of  the  thread  may  be 
constructed  by  passing  planes  of  rays  tangent  to  this  surface,  and 
joining  the  points  of  contact  by  a  line,  Art.  (238). 

To  find  the  point  of  this  curve  on  the  inner  helix,,  pass  a  plane 
tangent  to  the  lower  helicoid  at  F;  gd  is  the  horizontal  trace  of 
this  plane,  the  distance  fd  being  equal  to  the  rectified  arc  fee. 
Art.  (139). 

It  is  a  property  of  the  helicoid  that  all  tangent  planes  to  the 
surface,  at  points  of  the  same  helix,  make  the  same  angle  with  the 
axis*  or  with  the  horizontal  plane,  when  taken  perpendicular  to 
this  axis.  For  each  of  these  planes  is  determined  by  a  tangent  to 
the  helix,  and  the  rectilinear  element  through  the  point  of  contact; 
and  these  lines  at  all  points  of  the  helix  make  the  same  angle 
with  the  axis,  Arts.  (69)  and  (92),  and  with  each  other. 

Hence,  if  through  the  axis  we  pass  a  plane  perpendicular  to 
the  tangent  plane  at  F,  it  will  cut  from  this  plane  a  right  line,  in- 
tersecting oc  at  o",  horizontally  projected  in  cm,  and  making  with 
H  the  same  angle  as  that  made  by  a  plane  tangent  at  any  point 
of  the  helix,  and  with  the  rectilinear  element  passing  through  ihe 
point  of  contact  an  angle,  which  is  the  same  for  all  these  planes; 
hence  the  angle  iof  m\\  be  the  same  for  all  these  planes. 

If  this  line  be  now  revolved  about  oc\  it  will  generate  a  cone 
whose  rectilinear  elements  make  the  same  angle  with  H  as  the  tan- 
gent plane.  If  a  plane  of  rays  be  passed  tangent  to  this  cone,  it 
will  be  parallel  to  the  plane  of  rays  tangent  to  the  surface  at  a 
point  on  the  inner  helix,  hi,  tangent  to  the  circle  generated  by 
oiy  is  the  horizontal  trace  of  this  plane,  Art.  (131),  and  ol  is  the 


8HADES  AND  SHADOWS.  155 

horizontal  -projection  of  the  line  cut  from  the  parallel  tangent  plane 
of  rays  by  a  perpendicular  plane  through  oc'.  If,  then,  we  make 
the  angle  lox  equal  to  iof  ox  will  be  the  horizontal  projection  of 
the  element  containing  the  required  point  of  contact,  and  X  will 
be  the  point. 

In  the  same  way,  the  point  Y  on  the  outer  helix  may  be  ob- 
tained, by  first  passing  a  plane  tangent  to  the  lower  surface  at  M. 
eg  is  its  horizontal  trace,  mq  being  equal  to  the  rectified  semi-cir- 
cumference myn,  and  eop  the  constant  angle.  A  tangent  from  t  to 
the  circle  generated  by  op  (so  nearly  coincident  with  ecf  that  it  is 
not  drawn)  is  the  horizontal  trace  of  the  parallel  plane  of  rays,  and 
oy,  making  with  the  right  line  joining  o  and  the  point  of  contact 
of  this  tangent,  an  angle  equal  to  eop,  the  horizontal  projection  of 
the  element  containing  the  point  of  contact,  and  Y  the  required 
point. 

Intermediate  points  of  the  curve  of  shade  may  be  determined  by 
constructing,  as  above,  the  points  of  contact  of  tangent  planes  of 
rays  on  intermediate  helices.  Thus,  pass  a  plane  tungent  to  the 
lower  helicoid  at  W,  midway  between  F  and  N ;  wbz  is  the  hori- 
zontal projection  of  the  helix  passing  through  this  point;  fb"  the 
horizontal  projection  of  the  intersection  of  the  tangent  plane  with 
the  horizontal  plane  through  /',  10b"  being  equal  to  the  rectified 
arc  wb.  Since  this  intersection  is  parallel  to  the  horizontal  trace 
of  the  tangent  plane,  ou  will  be  the  horizontal  projection  of  the 
line  cut  from  the  tangent  plane  by  the  perpendicular  plane 
through  oc\  and/oa  the  constant  angle,  uh  is  the  horizontal  pro- 
jection of  the  circle  cut  from  the  auxiliary  cone  by  the  horizontal 
plane  through  fm',  and  ts  the  horizontal  projection  of  the  inter- 
section of  the  tangent  plane  of  rays  with  the  same  plane ;  oz  making 
with  the  right  line  joining  o  and  s,  an  angle  equal  to  foil,  is  then 
the  horizontal  projection  of  the  element  containing  the  point  ot 
contact,  o"V  Art.  (92)  its  vertical  projection,  and  Z  the  required 
point. 

As  the  surface,  of  which  e'm"ri'f"  is  the  vertical  projection,  is 
in  all  respects  identical  with  that  of  which  em'u'f  is  the  projec- 


156  DESCRIPTIVE    GEOMETRY. 

tion,  y"x"  will  be  the  vertical  projection  of  the  curve  of  shade  on 
the  first  surface,  its  horizontal  projection  being  yx. 

To  construct  the  shadow  cast  by  this  curve  of  shade  on  the  sur- 
face of  the  thread  below  it,  construct  the  shadow  of  (xy,  x"y")  on 
the  horizontal  plane  through  n'e',  under  the  supposition  that  the 
rays  are  unobstructed.  x'"y'"  is  the  horizontal  projection  of  this 
shadow,  Art.  (244.)  Then  assume  any  element,  as  (ao,  a'o;),  on 
which  it  is  supposed  the  shadow  will  fall,  and  construct  its  shadow 
on  the  same  'plane.  h"r"  is  the  horizontal  projection  of  this 
shadow.  Through  the  point  in  which' these  shadows  intersect, 
horizontally  projected  at  r" ,  draw  a  ray  of  light,  intersecting  the 
element  in  R,  which  is  the  shadow  of  the  curve  on  the  element, 
Art.  (244  )  The  curve  evidently  begins  at  (.r,  x")  and  is  vertically 
projected  in  x"r'. 

The  shadow  cast  by  the  helix  {run,  m"n")  on  the  surface  of  the 
thread  may  be  constructed  in  the  same  way  by  first  constructing 
its  shadow  on  the  same  horizontal  plane  as  above,  and  the  shadow 
of  an  assumed  element  (od,  o2(5')  on  the  same  plane,  and  from  their 
point  of  intersection  horizontally  projected  at  6",  drawing  a  ray ; 
(6,  6')  will  be  a  point  of  the  required  shadow. 

The  curve  of  shadow  on  the  surface,  whose  vertical  projection 
is  enf ,  will  be  equal  to  the  curve  whose  vertical  projection  is 
x"rf5\  and  may  be  drawn  as  in  the  figure. 


259.  The  brilliant  point  on  the  upper  surface  of  the  thread 
may  be  constructed  by  bisecting  the  angle  formed  by  a  ray  of 
light  and  a  line  drawn  to  the  point  of  sight,  Art.  (247),  and  then 
passing  a  plane  perpendicular  to  the  bisecting  line  and  tangent  tc 
the  surface,  as  in  Art.  (140),  and  finding  its  point  of  contact. 


PART  IV. 


LINEAR    PERSPECTIVE. 


PRELIMINARY    DEFINITIONS    AND    PRINCIPLES. 

260.  It  has  been  observed,  Art.  (204),  that  the  orthographic 
projections  can  never  present  to  the  eye  of  an  observer  a  perfectly 
natural  appearance,  and  hence  this  mode  of  representation  is  used 
onl\~  in  drawings  made  for  the  development  of  the  principles,  and 
the  solution  of  problems  in  Descriptive  Geometry,  and  for  the  pur- 
poses of  mechanical  or  architectural  constructions. 

Whenever  an  accurate  picture  of  an  object  is  desired,  the  sceno- 
graphic  method  must  be  used,  and  the  position  of  the  point  of 
sight  chosen,  as  indicated  in  Art.  (204). 


261.  That  application  of  the  principles  of  Descriptive  Geometry 
which  has  for  its  object  the  accurate  representation,  vpon  a  single 
plane,  of  the  details  of  the  form  and  the  principal  lines  c/  a  body, 
is  called  Linear  Perspective. 

The  art  by  which  a  proper  coloring  is  given  to  all  parts  <>f  the 
representation,  is  called  Aerial  Perspective.  This,  properly,  form 
no  part  of  a  mathematical  treatise,  and  is  therefore  left  entirely  t( 
the  taste  and  skill  of  the  artist. 


158  DESCRIPTIVE   GEOMETRY. 

262.  The  plane  upon  which  the  representation  of  the  body  is 
made  is  called  the  plane  of  the  picture  ;  and  a  point  is  represented 
upon  it,  as  in  all  other  cases,  Art.  (3),  by  drawing  through  the 
point  and  the  point  of  sight  a  right  line.  The  point  in  which  it 
pierces  the  plane  of  the  picture  is  the  perspective  of  the  given 
point. 

These  projecting  lines  are  called  visual  rays,  and  when  drawn 
to  the  points  of  any  curve,  form  a  visual  cone,  Art.  (11). 

Any  plane,  passing  through  the  point  of  sight,  is  made  up  of 
visual  rays,  and  is  called  a  visual  plane. 

The  plane  of  the  picture  is  usually  taken  between  the  object  to 
be  represented  and  the  point  of  sight,  in  order  that  the  drawing 
may  be  of  smaller  dimensions  than  the  object.  It  is  also  taken 
vertical,  as  in  this  position  it  will,  in  general,  be  parallel  to  many 
principal  lines  of  the  object.  It  may  thus  be  used  as  the  vertical 
plane  of  projection,  and  will  be  referred  to  as  the  plane  V, ' 
Art.  (24). 

The  orthographic  projection  of  the  point  of  sight,  on  the  plane 
of  the  picture,  is  called  the  principal  point  of  the  picture  ;  and  a 
horizontal  line  through  this  point  and  in  the  plane  of  the  picture, 
is  the  horizon  of  the  picture. 


PERSPECTIVES  OF  POINTS  AND  RIGHT  LINES.       VANISHING    POINTS    OF 
RIGHT    LINES. 

263.  Let  M,  Fig.  113,  be  any  point  in  space,  AB  the  horizontal 
trace  of  the  plane  of  the  picture,  or  ground  line,  and  S  the  point 
of  sight. 

The  visual  ray  SM,  through  M,  pierces  the  plane  of  the  picture 
V,  in  mx  Art.  (40),  which  is  the  perspective  of  the  point  M. 


264.  The  indefinite  perspective  of  a  right  line,  as  MN,  Fig.  114, 
may  be  drawn  by  finding  the  perspectives  of  any  two  of  its  points 


LINEAR   PERSPECTIVE.  159 

as  in  the  preceding  article,  and  joining  them  by  a  right  line ;  or 
by  passing  through  the  line  a  visual  plane,  and  constructing  its 
trace  on  the  plane  of  the  picture. 

The  point  n',  in  which  this  line  pierces  V,  is  one  point  of  this 
trace,  Art.  (30).  If  through  S  a  line  SP  be  drawn  parallel  to 
MN,  it  will  lie  in  the  visual  plane,  and  pierce  V  at  p',  a  second 
point  of  the  trace,  and  p'n!  will  be  the  perspective  of  the  line. 

The  point  p'  is  called  the  vanishing  point  of  the  line  MN. 
Hence,  to  construct  the  perspective  of  any  right  line,  join  its 
vanishing  point  with  the  point  where  the  line  pierces  the  plane  of  the 
picture,  by  a  right  line. 


265.  The  vanishing  point  of  any  right  line  is  the  point  in  which 
a  line,  parallel  to  it  through  the  point  of  sight,  pierces  the  plane  of 
the  picture. 

This  point,  as  seen  in  the  preceding  article,  is  always  one  point 
of  the  perspective  of  the  line  ;  and  since  the  auxiliary  line  intersects 
the  given  line  only  at  an  infinite  distance,  this  point  is  the  perspec- 
tive of  that  point  of  the  given  line  which  is  at  an  infinite  dis- 
tance. 

The  principal  point  is  evidently  the  vanishing  point  of  all  per- 
pendiculars to  the  plane  of  the  picture. 


266.  Any  horizontal  right  line  making  an  angle  of  45°  with  the 
plane  of  the  picture  is  a  diagonal. 

A  right  line  through  S,  Fig.  113,  parallel  to  a  diagonal,  pierces 
V  in  the  horizon,  and  at  a  distance  from  the  principal  point  s\ 
equal  to  the  distance  of  S  from  the  plane  of  the  picture.  Since 
two  such  lines  can  be  drawn,  one  on  each  side  of  the  perpendicular 
Ss',  there  will  be  two  vanishing  points  of  diagonals,  one  for  those 
which  incline  to  the  right,  and  another  for  those  which  incline  to 
the  left. 

These  points  are  also  called  points  of  distance,  since  when  as- 


160  DESCRIPTIVE    GEOMETRY. 

suraed  or  fixed,  the  distance  of  the  point  of  sight  from  the  plane 
of  the  picture  is  determined. 


267.  If  a  plane  be  passed  through  the  point  of  sight  parallel  to  a 
given  plane,  it  will  contain  all  right  lines  drawn  through  this  point 
parallel  to  lines  of  the  plane.  Its  trace  on  the  plane  of  the  pic- 
ture will  therefore  contain  the  vanishing  points  of  all  these  lines, 
and  of  lines  parallel  to  them.  This  trace  is  the  vanishing  line  oj 
the  plane. 

The  horizon  is  evidently  the  vanishing  line  of  all  horizontal 
planes. 

A  system  of  parallel  right  lines  will  have  a  common  vanishing 
point ;  since  the  line  through  the  point  of  sight  parallel  to  one  ia 
parallel  to  all ;  hence  their  perspectives  all  intersect  at  this  point, 
Art.  (265). 

If  the  lines  are  also  parallel  to  the  plane  of  the  picture,  the  aux- 
iliary line  will  be  parallel  to  it,  and  their  vanishing  point  at  an  in- 
finite distance.  Their  perspectives  will  then  be  parallel,  and  the 
perspective  of  each  line  parallel  to  itself,  and  it  will  be  necessary  to 
determine  only  one  point  of  its  perspective. 

If  a  right  line  pass  through  the  point  of  sight,  it  is  a  visual  ray, 
and  the  point  in  which  it  pierces  the  plane  of  the  picture  is  its 
perspective. 

208.  If  a  point  be  on  a  line,  its  perspective  will  be  on  the  per- 
spective of  the  line.  If  then  through  any  point  two  right  lines 
t)e  drawn,  and  their  perspectives  found  as  in  Art.  (264),  the  in- 
tersection of  these  perspectives  will  be  the  perspective  of  the  given 
point.  Hence,  in  practice,  the  perspective  of  a  point  is  constructed 
by  drawing  the  perspectives  of  a  diagonal  and  'perpendicular,  which, 
in  space,  pass  through  the  point,  and  finding  the  intersection  of 
these  perspectives. 

Thus  let  s',  Fig.  113,  be  the  principal  point,  s'd{  the  horizon 
d{  one  point  of  distance,  andM  any  poirt  in  space. 


LINEAR    PERSPECTIVE.  161 

MN  is  a  diagonal  through  the  point,  its  vertical  projection  m'n 
being  parallel  to  AB,  Art.  (14).  It  pierces  V  at  n',  vanishes  at  d, 
and  n'dx  is  its  perspective,  Art.  (264).  The  perpendicular  through 
M  pierces  V  at  m\  vanishes  at  s',  and  mV  is  its  perspective. 
These  perspectives  intersect  at  to,  the  perspective  of  M. 

If  the  point  M  should  be  in  the  horizontal  plane,  the  diagonal 
and  perpendicular  would  pierce  V  in  AB. 

When  the  given  point  is  near  the  horizontal  plane  through  the 
point  of  sight,  the  perspectives  of  the  diagonal  and  perpendicular 
through  it  are  so  nearly  parallel  that  it  is  difficult  to  mark  accu- 
rately their  point  of  intersection.  In  this  case,  find  the  perspec- 
tive of  a  vertical  line  through  the  given  point ;  its  intersection  with 
the  perspective  of  the  diagonal  or  perpendicular  will  be  the  re- 
quired point.  Thus  in  Fig.  113,  find  m2,  the  perspective  of  m  the 
foot  of  a  vertical  line  through  M,  and  draw  mjni}  intersecting 
m V  in  mx  the  perspective  of  M. 


269.  Since  the  object  to  be  represented  is  usually  behind  the 
plane  of  the  picture,  Art.  (262),  and  is  given  by  its  orthographic 
projections,  these  projections  when  made  as  in  Art.  (4),  will 
occupy  the  same  part  of  the  drawing  as  the  perspective,  and  cause 
confusion.  To  avoid  this,  in  some  degree,  that  portion  of  the 
horizontal  plane  which  is  occupied  by  the  horizontal  projection  of 
the  object,  is  revolved  about  AB  180°,  until  it  comes  in  front  of 
the  plane  of  the  picture.  Thus,  in  Fig.  113,  m  comes  to  m",  and 
the  horizontal  projection  of  the  diagonal  MN  to  the  position  m"n, 
its  vertical  projection  being  in  its  primitive  position,  and  the  point 
n'  being  found  as  before,  Art.  (27).  It  will  be  observed  that  the 
horizontal  projection  of  each  diagonal,  in  this  new  position,  lies 
in  a  direction  contrary  to  that  of  its  true  position.  The  vanishing 
point  will  be  determined  by  its  true  direction,  Art.  (266). 
11 


162  DESCRIPTIVE   GEOMETRY. 


PERSPECTIVES    OF    CURVES. 

270.  The  perspective  of  any  curve  may  be  found  by  joining  ita 
points  with  the  point  of  sight  by  visual  rays,  thus  forming  a  visual 
cone.  The  intersection  of  this  cone  by  the  plane  of  the  picture 
will  be  the  required  perspective  :  Or  the  perspectives  of  its  points 
may  be  found  as  in  Art.  (268),  and  joined  by  a  line. 

If  the  curve  be  of  single  curvature,  Art.  (58),  and  its  plane  pass 
through  the  point  of  sight,  its  perspective  will  be  a  right  line. 


271.  If  two  lines  are  tangent  in  space)  their  perspectives  will  be 
tangent  For  the  visual  cones,  by  which  their  perspectives  are 
determined,  will  be  tangent  along  the  common  rectilinear  element 
passing  through  the  point  of  contact  of  the  lines;  hence  their 
intersection  by  the  plane  of  the  picture  will  be  tangent  at  the 
perspective  of  this  point  of  contact. 


272.  If  the  circumference  of  a  circle  be  parallel  to  the  plane  of 
the  picture,  its  perspective  will  be  the  circumference  of  a  circle 
whose  centre  is  the  perspective  of  the  given  centre ;  also,  if  it  be 
bo  situated  that  the  plane  of  the  picture  makes  in  its  visual  cone  a 
sub-contrary  section,  Art.  (206). 

In  all  other  cases,  when  behind  the  plane  of  the  picture,  and  its- 
plane  does  not  pass  through  the  point  of  sight,  its  perspective  will 
be  aw  ellipse. 

It  is  evident  that  if  the  condition  be  not  imposed  that  the  circle 
shall  be  behind  the  plane  of  the  picture,  this  plane  may  be  so 
taken  a^s  to  intersect  the  visual  cone  in  any  of  the  conic  sections. 


273.  Let  mlo*,  Fig.  115,  be  a  circle  in  the  horizontal  plane,  re- 
volved as  in  Art.  (269),  and  let  s'  be  the  principal  point,  the 


LINEAR   PERSPECTIVE.  103 

point  of  sight  being  taken  in  a  plane  through  the  centre  and  per- 
pendicular to  AB,  and  let  dx  and  dt  be  the  points  of  distance. 

o,  is  the  perspective  of  o,  and  mx  of  w,  Art.  (268),  and  oxmx  of 
the  diameter  om.  The  perspectives  of  the  two  tangents  at  o  and  m 
will  be  tangent  to  the  perspective  of  the  circle  at  ox  and  mu  Art. 
(271);  and  since  the  tangents  are  parallel  to  AB,  their  perspec- 
tives will  be  parallel  to  AB,  Art.  (267),  or  perpendicular  to  oxmx ; 
hence  oxmi  is  an  axis  of  the  ellipse.  Through  its  middle  point  et 
draw  dxex ;  it  is  the  perspective  of  the  diagonal  fe,  and  ex  is  the 
perspective  of  e,  and  uiz]  of  the  parallel  chord  uz1  u's'  and  z,sf 
being  the  perspectives  of  the  perpendiculars  through  u  and  2,  and 
tt,z,  is  the  other  axis  of  the  ellipse.  On  these  two  axes  the  ellipse 
can  be  described  as  in  Art.  (59). 

The  perspectives  of  the  two  perpendiculars  IV  and  Jck'  are  tan- 
gent to  the  ellipse  at  £,  and  hx. 

If  the  point  of  sight  be  taken  in  any  other  position,  the  per- 
spective of  the  circle,  Fig.  116,  may  be  determined  by  points,  as 
in  Art.  (270),  and  the  curve  drawn  through  these  points  tangent 
to  the  perspectives  of  the  tangents  at  o,  m,  I  and  k.  The  line 
OjWi,  will  be  a  conjugate  diameter  to  the  line  uxzx,  since  the  tan- 
gents at  o,  and  mx  are  parallel  to  AB  and  to  uxzx* 


LINE    OF    APPARENT    CONTOUR. 

274.  If  a  body,  bounded  by  a  curved  surface,  be  enveloped  by 
a  tangent  visual  cone,  the  line  of  contact  of  this  cone  will  be  the 
outer  line  of  the  body,  as  seen  from  the  point  of  sight,  and  is  the 
line  of  apparent  contour  of  the  body. 

The  perspective  of  this  line,  Art.  (270),  is  evidently  the  bound- 
ing line  of  the  picture  or  drawing,  and  is  a  principal  line  of  the 
perspective. 

When  the  body  is  bounded  by  plane  surfaces,  the  visual  cone 
will  be  made  up  of  visual  planes ;  the  line  of  apparent  contour 
will  not  be  a  line  of  mathematical  contact,  but  will  still  be  the 


1G4  DESCRIPTIVE    GEOMETRY. 

outer  line  of  the  body  as  seen,  and  will  be  made  up  of  right 
lines. 

When  the  body  is  irregular,  or  composed  of  broken  surfaces, 
the  line  of  contour  may  be  composed  partly  of  strict  lines  of  con- 
tact, either  straight  or  curved,  and  partly  of  lines  not  of  contact, 
but  still  the  outer  lines  of  the  body  as  seen.  In  all  cases  their 
perspectives  will  form  the  boundary  of  the  picture. 

To  construct  the  perspective  of  any  body,  we  must  then  deter- 
mine the  perspective  of  its  apparent  contour,  and  of  such  other 
principal  lines  as  will  aid  in  indicating  its  true  form. 


275.  If  a  line  of  the  surface  intersect  the  line  of  contact  of  the 
visual  cone,  the  perspective  of  this  line  will  be  tangent  to  the  per- 
spective of  the  line  of  contact.  For  at  the  point  of  intersection 
draw  a  tangent  to  each  of  the  two  lines;  these  tangents  lie  in  the 
tangent  plane  to  the  surface,  and  this  plane  is  also  a  visual  plane 
tangent  to  the  cone.  Its  trace  on  the  plane  of  the  picture  is  the 
perspective  of  both  tangents,  and  tangent  to  the  perspectives  oi 
both  lines  at  a  common  point.  They  are  therefore  tangent  to  each 
other. 


VANISHING  POINTS  OF  RAYS  OF  LIGHT  AND  OF  PROJECTIONS  OF  RAYS. 

270.  Since  the  rays  of  light  are  parallel  they  will  have  a  com- 
mon vanishing  point,  which  may  always  be  determined  by  draw- 
ing a  visual  ray  of  light,  and  finding  the  point  in  which  it  pierces 
V,  Art.  (265).  In  the  construction  of  a  drawing  or  picture,  the 
direction  of  the  light  is  usually  chosen  by  the  draughtsman  or 
artist.  This  is  done  by  assuming  the  vanishing  point  of  rays  at 
once,  as  the  direction  of  the  light  is  thus  completely  determined, 
Thus  let  r„  Fig.  117,  be  the  vanishing  point  of  rays,  S  being  the 
point  of  sight,  sr  will  be  the  horizontal  projection  of  the  ray,  and 
*V,  'ts  vertical  projection. 


LINKAR   PERSPECTIVE.  165 

277.  The  horizontal  projections  of  rays  of  light  being  parallel 
and  horizontal  lines,  must  vanish  in  the  horizon,  Art.  (267)  ;  and 
since  a  line  through  the  point  of  sight  parallel  to  the  horizontal 
projection  of  a  ray,  must  be  in  the  same  vertical  plane  with  a  ray 
through  the  same  point,  it  must  pierce  the  plane  of  the  picture  in 
the  vertical  trace  of  this  plane,  that  is,  in  the  line  rxr'  at  r'. 
Hence,  having  assumed  the  vanishing  point  of  rays,  through  it 
draw  a  right  line  perpendicular  to  the  horizon ;  it  will  intersect  it 
in  the  vanishing  point  of  the  horizontal  projections  of  rays. 


278.  The  orthographic  projections  of  rays  on  all  planes  perpen- 
dicular to  AB,  as  the  plane  tTt',  are  parallel.  The  line  s's  is  the 
vanishing  line  of  these  planes,  Art.  (267),  and  must  therefore 
contain  the  vanishing  point  of  these  projections.  This  point  must 
also  be  in  the  trace  of  a  plane  of  rays  through  S,  perpendicular  to 
these  side  planes.  r,r2,  parallel  to  AB,  is  this  trace ;  hence  r2  is 
the  vanishing  point  of  the  projections  of  rays  on  side  planes. 


279.  The  orthographic  projections  of  rays  on  the  plane  of  the 
picture,  or  on  planes  parallel  to  it,  are  parallel ;  and  being  in,  or 
parallel  to,  the  plane  of  the  picture,  will  be  parallel  in  perspective, 
Art.  (267),  and  all  parallel  to  r{s',  the  projection  of  the  ray 
through  S  on-V. 


PERSPECTIVES    OF    THE    SHADOWS    OF    POINTS    AND    RIGHT   LINES    ON 
PLANES. 

280.  Since  the  shadow  of  a  point  on  a  plane  is  the  point  in  which 
a  ray  of  light  through  the  point  pierces  the  plane,  Art.  (241),  it 
must  be  the  intersection  of  the  ray  with  its  orthographic  projection 
on  the  plane.  Hence,  to  construct  the  perspective  of  the  shadow  of 
a  point  on  any  plane,  through  the  perspective  of  the  point  draw 


166  DESCRIPTIVE   GEOMETRY. 

the  perspective  of  a  ray.  and  through  the  perspective  of  the  pro- 
jection of  the  point  on  the  plane,  draw  the  perspective  of  the  pro- 
jection of  the  ray ;  the  intersection  of  these  two  lines  will  be  the 
perspective  of  the  required  shadow,  Art.  (268). 

If  the  shadow  be  on  the  horizontal  plane,  join  the  perspective  of 
the  point  with  the  vanishing  point  of  rays  ?*,,  Fig.  117,  and  the 
perspective  of  the  horizontal  projection  of  the  point  with  the  van- 
ishing point  of  horizontal  projections  /;  the  intersection  of  these 
two  lines  will  be  the  perspective  of  the  shadow  of  the  given 
point. 

If  the  shadow  be  on  any  side  plane,  join  the  perspective  of  the 
projection  of  the  point  on  this  plane  with  the  vanishing  point  of 
the  projections  of  rays  on  side  planes  r2;  the  intersection  of  this 
line  with  the  perspective  of  the  ray,  will  be  the  perspective  of  the 
shadow. 

If  the  shadow  be  on  any  plane  parallel  to  V,  through  the  per- 
spective of  the  projection  of  the  point  on  this  plane  draw  a  line 
parallel  to  the  projection  of  the  ray  on  the  plane  of  the  picture 
s'r, ;  its  intersection  with  the  perspective  of  the  ray  will  be  the 
perspective  of  the  shadow. 

If  the  shadow  be  on  any  vertical  plane,  draw  through  the  per- 
spective of  the  horizontal  projection  of  the  point,  a  right  line  to 
the  vanishing  point  of  horizontal  projections.  It  will  be  the  per- 
spective of  the  horizontal  trace  of  a  vertical  plane  of  rays  through 
the  point.  At  the  point  where  it  intersects  the  perspective  of  the 
horizontal  trace  of  the  given  plane  erect  a  vertical  line ;  it  will  be 
the  perspective  of  the  intersection  of  the  plane  of  rays  with  the 
given  plane.  The  point  where  this  intersects  the  perspective  of 
the  ray  through  the  given  point,  will  be  the  perspective  of  the 
shadow. 


281.  The  perspective  of  the  indefinite  shadow  cast  by  a  right 
line  on  a  plane,  may  be  constructed  by  finding  the  perspective  of- 
the  point  in  which  the  line  pierces  the  plane,  aud  joining  it  with 


LINEAR   PERSPECTIVE.  167 

the  perspective  of  the  shadow  cast  by  any  other  point  of  the  right 
line ;  or  by  joining  the  perspectives  of  the  shadows  of  any  two 
points  of  the  line  by  a.  right  line. 

If  the  line  be  of  definite  length  join  the  perspectives  of  the 
shadows  of  its  two  extremities  by  a  right  line. 


PRACTICAL    PROBLEMS, 

282.  Problem  70.  To  construct  the  perspective  of  an  upright  rect- 
angular pillar,  with  its  shade  and  shadow  on  the  horizontal  plane. 

Let  Imno,  Fig.  118,  be  the  lower  base  of  the  pillar  in  the  hori- 
eontal  plane,  revolved  as  in  Art.  (269),  and  p'q'  the  vertical  pro- 
jection of  the  upper  base,  s'  the  principal  point,  dt  one  of  the 
points  of  distance,,  r{  the  vanishing  point  of  rays,  and  r'  of  hori- 
zontal projections  of  rays.  These  important  points  will  be  .thus 
represented  in  all  the  following  problems. 

Since  ml  and  no  are  perpendicular  to  V,  they  vanish  at  sf,  Art. 
(265),  and  m's'  and  n's'  are  their  indefinite  perspectives  ;  id2  is 
the  perspective  of  the  diagonal  ni,  and  w,  of  the  point  n,  Art, 
(268).  »,?»,',  parallel  to  AB,  is  the  perspective  of  mn,  Art.  (267); 
ox  of  the  point  o  ;  n{o{  of  the  edge  no,  Art.  (264);  o^,  of  ol,  and 
mxl{  of  ml. 

The  edges  of  the  upper  base,  which  are  horizontally  projected- 
in  on  and  Im,  pierce  V  at  p'  and  q  ,  and  vanish  at  s' ;  hence  p's' 
and  q's'  are. their  indefinite  perspectives.  The  diagonal  of  the 
upper  base,  horizontally  projected  in  ni,  pierces  V  at  i',  i'd2  is  its 
perspective,  and  px  the  perspective  of  the  vertex  of  the  upper  base 
horizontally  projected  in  n,  and  vx  of  that  horizontally  projected  at 
1.  The  vertical  edges  which  pierce  H  in  m,  n,  o,  and  I,  are  parallel 
to  Y,  and  their  perspectives  parallel  to  themselves ;  hence  mxq^ 
w,^„  oxux,  lxvx  are  their  perspectives  terminating  in  the  points  qup„ 
a„  Vj,  and  q\P\Urv{  is  the  perspective  of  the  upper  base. 

The  face  of  which  n^o^uxp^  is  the  perspective  is  in  the  shade,  an 
therefore  is  darkened  in  the  drawing. 


168  DESCRIPTIVE   GEOMETRY. 

The  shadow  on  II,  of  the  edge  represented  by  nxpx  is  rc,^?2,  Art 
(281).  The  shadow  of  the  edge  represented  by  pxux  is  parallel  to 
the  line  itself,  and  therefore  perpendicular  to  V,  and  vanishes  at 
*'.  It  is  limited  at  />2  and  w2,  Art.  (281).  w2v2,  parallel  to  AB, 
is  the  perspective  of  the  shadow  of  the  edge  represented  by  uxv  , 
and  lxv2  of  that  represented  by  lxvx. 

That  part  of  the  drawing  within  the  line  n^p^v^  is  darkened,  be- 
ing the  perspective  of  that  part  of  the  shadow  on  H  which  is  seen; 


283.  Problem  71.  To  construct  the  perspective  of  a  square  py- 
ramid with  its  pedestal,  and  the  perspective  of  its  shadow. 

Let  mnol.  Fig.  119,  be  the  base  of  the  pedestal  revolved,  as  in 
Art.  (269),  and  mnn'vri  its  front  face  in  the  plane  of  the  picture, 
pqtu  the  horizontal,  and  p'u'  the  vertical  projection  of  the  base  of 
the  pyramid,  and  0  its  vertex. 

The  face  mnit'm'  being  in  the  plane  of  the  picture  is  its  own 
perspective.  The  four  edges  of  the  pedestal,  which  are  perpen- 
dicular to  V,  pierce  it  in  m,  w,  n  and  m!  and  vanish  at  *'.  The 
two  diagonals  mo  and  (rao,  m'n')  pierce  V  at  m  and  m'  and  vanish 
at  dv  The  diagonals  nl  and  (w/,  n'm')  pierce  V  at  rt  and  ri  and 
vanish  at  d.2.  Hence  o,  is  the  perspective  of  o,  Art.  (268)  ;  kt  of 
(on') ;  lx  of  / ;  and  A,  of  (lm') ;  and  the  perspective  of  the  pedestal 
is  drawn  as  in  Art.  (282). 

The  two  perpendiculars  (at,  u')  and  (pq,  p')  pierce  V  at  u'  and 
pr ;  u's'  and  p's'  are  their  perspectives  intersecting  m'dx  and  n'd^ 
in  «„  <7„  p{  and  tx  and  uipxqitl  is  the  perspective  of  the  base  of  the 
pyramid. 

The  perpendicular  through  C  pierces  V  at  c\  and  a  diagonal 
through  the  same  point  at  h!  and  c's'  and  h'dx  are  their  perspec- 
tives, and  cx  the  perspective  of  C;  and  cxu}1  cspx,  cxq^  and  cltl  are 
the  perspectives  of  the  edges  of  the  pyramid. 

nn2  is  the  perspective  of  the  shadow  of  nn  on  H ;  n2&2  of  the 
shadow  of  the  edge  represented  by  n'kx  Art.  (265).  ix  is  the  per- 
spective of  c,  the  horizontal  projection  of  the  vertex  •  ixr'  of  the 


LINEAR   PERSPECTIVE.  169 

horizontal  projection  of  a  ray  through  C,  and  cxr{  the  perspective 
of  the  ray ;  hence  c2  is  the  perspective  of  the  shadow  of  C  on  H, 
Art.  (280).  e}  is  the  perspective  of  the  projection  of  C  on  the 
upper  base  of  the  pedestal,  exr'  of  the  projection  of  a  ray  on  this 
plane,  and  c8  of  the  shadow  cast  by  C  on  this  plane ;  and  p{c3  the 
perspective  of  the  shadow  cast  by  the  edge  CP  on  this  plane, 
Art  (281).  This  shadow  passes  from  the  upper  base  at  a  point 
of  which  r,  is  the  perspective;  #,?•,  is  the  perspective  of  a  ray 
through  this  point,  intersecting  n.2s'  at  $*  the  perspective  of  the 
shadow  cast  by  one  point  of  the  edge  CP  on  H,  and  x.2c2  is  the 
perspective  of  the  shadow. 

ttc3  is  the  perspective  of  the  shadow  cast  by  the  edge  CT  on 
the  upper  base  of  the  pedestal ;  yx  of  the  point  at  which  it  parses 
from  this  base ;  y.2  of  the  shadow  of  this  point  on  H,  and  y^c2  of 
the  shadow  of  the  edge  CT. 

The  face  represented  by  cxrp,q[  is  in  the  shade,  as  also  that  re- 
presented by  nn'kxo^  and  both  are  darkened  on  the  drawing. 

2>\X-Jcyxtx  bounds  the  darkened  part  on  the  perspective  of  the 
upper  base,  and  nn2 . .  c.^2  .  .  lx  that  on  H. 


284.  Problem  72.  To  construct  the  perspective  of  a  cylindrical 
column  with  its  square  pedestal  and  abacus,  and  also  the  shade  of 
the  column  and  shadow  of  the  abacus  on  the  column. 

Let  mngl  (Fig.  120)  be  the  horizontal  projection  of  both  ped- 
estal and  abacus,  Art.  (269) ;  mm'n'n  the  vertical  projection  of  the 
pedestal,  and  e'l'g'f  that  of  the  abacus ;  upqt  being  the  horizon- 
tal projection  of  the  column,  m'n'  the  vertical  projection  of  its 
lower,  and  e'f  of  its  upper  base;  the  plane  of  the  picture  being 
coincident  with  that  of  the  front  faces  of  the  pedestal  and  abacus. 

Let  the  point  of  sight  be  taken  as  in  Art  (273),  in  a  plane 
through  the  centres  of  the  upper  and  lower  bases  of  the  column 
and  perpendicular  to  AB. 

Construct  the  perspective  of  the  pedestal  as  in  Art  (282), 
mm'n'n  being  its  own  perspective,  and  m'lx  and  n'g{  the  perspec- 


170  DESCRIPTIVE    GEOMETRY. 

tives  of  those  edges  of  the  upper  face  of  the  pedestal  which  pierce 
V  at  rn'  and  n'. 

In  the  same  way  construct  the  perspective  of  the  abacus ;  e'l'g'f' 
being  its  own  perspective,  and  e'$'  and/ V  the  indefinite  perspec- 
tives of  the  edges  of  the  lower  face  which  pierce  V  at  e'  and/''. 

u\t\9\P\  is  the  perspective  of  the  lower  ba-e  of  the  column  de- 
termined as  in  Art.  (273);  uiqx  being  its  semi-conjugate  and  t]pi 
its  semi-tran verse  axis.  In  the  same  way  the  perspective  of  the 
upper  base  is  determined,  its  conjugate  axis  being  the  perspective 
of' that  diameter  which  pierces  V  at  ?/'',  horizontally  projected  in 
uq,  and  its  tranverse  axis  zvx  the  perspective  of  the  chord  corre- 
sponding to  tp. 

t(nx  and  pivl  tangent  to  these  ellipses,  Art.  (275),  and  perpen- 
dicular to  AB,  are  the  perspectives  of  the  elements  of  contour  of 
the  column,  and  complete  its  perspective,  Art.  (274). 

The  elements  of  shade  on  the  column  are  determined  by  two 
tangent  planes  of  rays,  Art.  (251),  and  since  these  planes  are  ver- 
tical, their  intersections  with  the  plane  of  the  upper  face  of  the 
pedestal  will  be  parallel  to  the  horizontal  projections  of  rays,  and 
therefore  vanish  at  ■/■'.  Since  these  intersections  are  also  tangent 
to  the  lower  circle  of  the  column,  their  perspectives  will  be  tan- 
gent to  the  ellipse  t^u  px.  Hence,  if  through  r'  two  tangents  be 
drawn  to  t  uxp ,  their  points  of  contact  will  be  points  of  the  per- 
spectives of  the  elements  of  shade.  atk2  is  the  perspective  of  the 
only  one  which  is  seen.  The  plane  of  rays  by  which  this  is  de- 
termined intersects  the  lower  edge  e'f  of  the  abacus  in  k]y  through 
which  draw  k{rim  It  intersects  uxkt  in  &,,  the  perspective  of  the 
point  of  shadow  cast  by  k},  and  limiting  the  perspective  of  the 
element  of  shade. 

Draw  r't^  ;  it  is  the  perspective  of  the  intersection  of  a  vertical 
plane  of  rays  with  the  upper  face  of  the  pedestal.  This  plane 
intersects  the  column  in  an  element  represented  by  txz ,  and  the 
edge  represented  by  e'a'  in  a  point  of  which  y,  is  the  perspective. 
Through  yx  draw  yxri  intersecting  f,z,  in  y2,  the  perspective  ol  the 
shadow  cast  on  the  column  by  the  point  Y,  and  in  the  same  way 


LINEAR    PERSPECTIVE.  171 

the  perspective  of  the  shadow  cast  by  any  point  of  the  same  edge 
may  be  determined. 

Through  m'  draw  m'r\  and  through  e'  draw  e'rx  intersecting 
the  perspective  of  the  element  cut  from  the  column  in  e.»  the  per- 
spective of  the  shadow  cast  by  e  on  the  column,  and  in  the  same  way 
the  perspective  of  the  shadow  cast  by  any  other  point  of  the  edge 
e'f  may  be  found,  as  w%  the  perspective  of  the  shadow  cast  by  w" . 

All  of  the  lower  face  of  the  abacus  is  in  the  shade,  and  is  dark- 
ened in  the  drawing.  The  line  y<2e<l...k.i  is  the  perspective  of  the 
line  of  shadow  on  the  column,  and  all  above  it  is  darkened,  as  all 
beyond  ajc2.  axr'  is  the  perspective  of  the  shadow  cast  by  the  ele- 
ment of  shade  on  the  upper  surface  of  the  pedestal,  and  the  part 
beyond  it  which  is  seen  is  therefore  darkened. 


285.  Problem  73.  To  construct  the  perspective  of  an  upright 
cylindrical  ring,  with  its  shade  and  shadow  on  its  interior  surfuce. 

Let  mngl,  Fig.  121,  be  the  horizontal  projection  of  the  outer 
cylinder  of  the  ring,  and  ezxh'  its  vertical  projection  ;  upqt  the 
horizontal,  and  olixwx  the  vertical  projection  of  the  inner  cylinder, 
the  plane  of  the  picture  being  coincident  with  the  plane  of  the 
front  face  of  the  ring. 

The  two  circles  ezxti  and  o{ixiox  being  in  the  plane  of  the  picture 
are  their  own  perspectives. 

To  construct  the  perspectives  of  the  back  circles  horizontally 
projected  in  Ig  and  tq,  draw  the  vertical  radius  (c,  c'h')  and  the 
two  vertical  tangents  at  Q  and  G.  A,c,  axqx,  and  b^g^  are  the  per- 
spectives of  these  lines,  Art,  (268),  and  c„  gu  and  g{  are  the  per- 
spectives of  the  points  C,  Q,  and  G.  With  c,  as  a  centre,  and  cxqx 
and  cxg,  as  radii,  describe  the  arcs  txt"  and  v{gxi  they  will  be  the 
perspectives  of  arcs  of  the  back  circles. 

sxyx  and  s{xx  are  the  perspectives  of  the  elements  of  contoui 
tangent  to  the  circles  eyKxK  and  vtgit  Art.  (275). 

The  element  of  shade  on  the  outer  cylinder  is  determined  by  a 
tangent  plane  of  rays.     This  plane  being  perpendicular  to  V,  its 


1 72  DESCRIPTIVE     GEOMETRY. 

vertical  trace  will   be  parallel   to  r,*',  Art.  (2*79),  and  tangent  tc 
yxzxh'  at  z ,  and  zivl  will  be  the  perspective  of  this  element. 

Points  of  the  shadow  cast  by  the  circle  o[i1ivl  on  the  interior 
cylinder  will  be  found  by  passing  planes  of  rays  perpendicular  to 
the  plane  of  the  picture.  Each  plane  will  intersect  the  circle  in 
a  point  which  casts  a  shadow  on  the  element  which  the  plane  cuts 
from  the  cylinder,  Art.  (252).  o,i,  parallel  to  r,#'  is  the  vertical 
trace  of  such  a  plane,  intersecting  the  circle  in  ob  and  the  cylin- 
der in  an  element  of  which  *,*'  is  the  perspective.  oxr,  is  the 
perspective  of  a  ray  through  o.,  and  o2  is  the  perspective  of  the 
shadow.     The  perspective  of  the  shadow  evidently  begins  at  k{. 


286.  Problem  74.  To  construct  the  perspective  of  an  inverted 
xrustum  of  a  cone  with  its  shade  and  shadow. 

Let  C,  Fig.  122,  be  the  vertex,  holm  the  horizontal,  Art.  (269), 
and  k'V  the  vertical  projection  of  the  upper  base,  ehgf  the  hori- 
zontal, and  e'g'  the  vertical  projection  of  the  lower  base,  and  kre' 
the  vertical  projection  of  one  of  the  extreme  elements. 

The  perspectives  of  the  bases  are  determined  as  in  Art.  (213), 
kxdy-pnx  of  the  upper,  and  elhiaif  that  of  the  lower  base. 

The  perspective  of  the  vertex  is  found  as  in  Art.  (268),  CZ 
being  the  diagonal  through  it,  piercing  V  at  z! ;  and  (co7  c')  the 
perpendicular  piercing  V  at  c',  z'd2  and  cV  are  their  perspectives 
intersecting  at  c,.  Through  cx  draw  the  two  tangents  cxwx  and 
c,y, ;  they  are  the  perspectives  of  the  elements  of  contour,  and 
with  the  curves  kxo'yxmx  and  exhxaxf  limit  the  perspective  of  the 
frustum. 

The  elements  of  shade  on  the  cone  are  determined  by  two 
tangent  planes  of  rays  intersecting  in  a  ray  of  light  passing 
through  the  vertex,  Art.  (131),  r,c,  is  the  perspective  of  this  ray. 
and  uxr'  the  perspective  of  its  horizontal  projection,  ux  being  the 
perspective  of  c;  hence  c2  is  the  perspective  of  the  point  in  which 
this  ray  pierces  H,  and  the  tangents  etix  and  c«px  are  the  perspec- 
tives of  the  horizontal  traces  of  the  tangent  planes,  and  ixnx  and 


LINEAR    PERSPECTIVE.  173 

pxqx  of  the  elements  of  shade,  Art.  (253),  the  former  only  being 
seen. 

That  part  of  the  circumference  of  the  upper  base  towards  the 
source  of  light,  and  between  the  points  of  which  nx  and  gx  are  the 
perspectives,  casts  its  shadow  on  the  interior  of  the  cone.  Points 
of  this  shadow  may  be  determined  by  intersecting  the  cone  by 
planes  of  rays  through  the  vertex.  Each  plane  intersects  the  cir- 
cumference in  a  point,  and  that  part  of  the  cone  opposite  the 
source  of  light  in  a  rectilinear  element.  A  ray  of  light  through 
th«  point  intersects  the  element  in  a  point  of  the  required 
shadow. 

c%ax  is  the  perspective  of  the  horizontal  trace  of  such  a  plane. 
The  plane  intersects  the  cone  in  two  elements  represented  by  axyx 
and  bxcx,  and  bx  is  the  perspective  of  the  point  in  which  the  plane 
intersects  the  circumference,  bxrx  the  perspective  of  a  ray  through 
this  point,  and  b.2  the  perspective  of  the  required  point  of 
shadow. 

The  curve  of  shadow  evidently  begins  at  the  points  of  which  nx 
and  qx  are  the  perspectives. 

The  perspective  of  the  lowest  point  of  this  shadow  will  be 
found  by  using  the  line  c^ie,,  us  this  is  the  perspective  of  the  trace 
of  that  plane  which  cuts  out  the  element  furthest  from  the  point 
casting  the  shadow. 

The  perspective  of  the  point  of  shadow  on  any  element  is  found 
by  using  the  line  drawn  through  c2  and  the  lower  extremity  of  the 
perspective  of  the  element.  Thus  the  point  of  tangency  bi7  Art. 
(275),  is  found  by  using  c.2ui  as  the  perspective  of  the  trace  of 
the  auxiliary  plane. 

<?!&,(/  is  i he  perspective  of  that  part  of  the  shadow  on  the  inte- 
rior which  is  seen,  and  is  therefore  darkened  in  the  drawing  as  is 
the  perspective  of  the  shade  n,ixa{y^. 

ix?i2  and  pxq.2  are  the  perspectives  of  the  shadows  cast  by  the 
elements  of  shade  on  H,  the  points  n2  and  q2  being  determined  bj 
w,n  and  qtrx. 

The  plane  determined  by  c.2ax  intersects  the  circumference  of  th 


174  DESCRIPTIVE    GEOMETRY. 

upper  base  in  a  point  of  which  y,  is  the  perspective,  ylri  is  the 
perspective  of  a  ray  through  this  point  piercing  H  at  a  point  of 
which  y2  is  the  perspective.  This  is  a  point  of  the  perspective  of 
the  curve  of  shadow  of  the  upper  circumference  on  H  ;  and  in  the 
same  way  any  number  of  points  may  be  determined. 

r^j  drawn  tangent  to  nxy{mx  will  also  be  tangent  to  ntftft,  since 
this  tangent  is  the  perspective  of  the  element  of  contour  of  the 
cylinder  of  rays  through  the  upper  circumference  by  which  its 
shadow  is  determined,  Art.  (275). 

The  curve  n$%q%  is  also  tangent  to  iAnt  and  pxq%  at  ^  and  q.2. 


287.  Problem  75.  To  construct  the  perspective  of  a  niche  with 
its  shadow  on  its  interior  surface. 

Let  the  niche  be  formed  by  a  semi-cylinder,  the  lower  base  of 
which  is  mlo,  Fig.  123,  and  the  upper  base  vertically  projected  in 
m'o\  and  the  quarter  sphere  vertically  projected  in  the  semicircle 
m'k'o',  its  lower  semicircle  being  coincident  with  the  upper  base 
of  the  cylinder,  and  the  plane  of  the  picture  so  taken  as  to  con- 
tain the  elements  mm'  and  oo'  and  the  front  circle  m'k'o'. 

These  elements  and  front  circle  being  in  the  plane  of  the  pic- 
ture will  be  their  own  perspectives. 

An  arc  of  an  ellipse,  mlxo,  is  the  perspective  of  the  lower  base 
of  the  cylindrical  part,  and  m'nxo'  that  of  the  upper  base,  these 
being  constructed  as  in  Art.  (273),  /,  being  the  perspective  of  I 
and  »,  of  the  corresponding  point  of  the  upper  base.  These  lines 
form  the  perspective  of  the  niche. 

The  lines  which  cast  shadows  on  the  interior  of  the  niche  are 
the  element  mm'  and  the  arc  rn'k'e'. 

The  shadow  cast  by  mm'  is  determined  by  passing  through  it  a 
vertical  plane  of  rays ;  mr'  is  the  perspective  of  its  trace,  Art. 
(277).  This  plane  cuts  from  the  cylinder  an  element  represented 
by/i'^i,  the  intersection  of  which  by  raV,  in  m.2  is  the  perspective 
of  the  shadow  cast  by  m'  on  the  element;  and  f%m%  is  the 
perspective   of  the   shadow  cast  by  m'f,  a  part  of  mm',  on  the 


LINEAR    PERSPECTIVE.  175 

interior  of  the  cylinder ;  mf  is  the  perspective  of  the  shadow  cast 
by  mf  on  H. 

Points  of  the  shadow  cast  by  the  front  circumference,  m'k'e\  on 
the  cylindrical  part  of  the  niche  may  be  determined  by  intersecting 
the  cylinder  by  vertical  planes  of  rays.  Each  plane  intellects  the 
cylinder  in  a  rectilinear  element  and  the  arc  in  a  point.  A  ray  of 
light  through  this  point  intersects  the  element  in  a  point  of  the 
curve  ot  shadow,  r'g  is  the  perspective  of  the  horizontal  trace  of 
such  a  plane.  It  intersects  m'k'e'  in  g',  and  the  cylinder  in  an 
element  represented  by  igt  and  g%  is  the  perspective  of  the  shadow 
cast  by  g'  on  the  element. 

The  shadow  cast  by  a  part  of  the  front  circumference  on  the 
spherical  part  of  the  niche  is  an  equal  arc  of  a  great  circle.  For 
this  shadow  is  determined  by  a  cylinder  of  rays  through  the  front 
circumference,  and  the  part  of  each  element  of  the  cylinder  be- 
tween the  point  casting  the  shadow  and  its  shadow  is  a  chord  of 
the  sphere,  and  all  these  chords  will  be  bisected  by  a  plane 
through  c'  perpendicular  to  them.  These  half  chords  may  be 
regarded,  one  set  as  the  ordi nates  of  the  arc  casting  the  shadow, 
and  the  other  as  the  ordinates  of  the  shadow,  and  since  the  cor- 
responding ordi  nates  of  the  two  arcs  are  equal,  and  in  all  respects 
like  situated,  the  two  arcs  will  be  equal.  Their  planes  evidently 
intersect  in  the  radius  c'e  perpendicular  to  the  axis  of  the  cylinder 
of  rays,  or  to  the  ray  of  light,  and  also  perpendicular  to  the  pro- 
jection s'r1  of  the  ray  of  light  on  the  plane  of  the  picture,  Art. 
(276). 

Points  of  this  shadow  may  be  constructed  by  intersecting  the 
quarter  sphere  by  planes  parallel  to  the  plane  of  the  picture. 
Each  plane  will  cut  from  the  quarter  sphere  a  semi-circumference. 
The  front  semi-circumference  will  cast  upon  this  plane  a  shadow 
parallel  and  equal  to  itself,  Art.  (246),  and  the  intersection  of  this 
with  the  semi-circumference  cut  from  the  sphere  will  be  a  point  of 
the  required  shadow,  Art.  (244). 

Draw  yxzx  parallel  to  mV,  it  may  be  taken  as  the  perspective 
of  the  intersection  of  an  auxiliary  plane  with   the   upper   base  oi 


176  DESCRIPTIVE    GEOMETRY. 

the  cylinder,  and  the  semi-circumference  yxxgx  is  the  perspective 
of  the  semi-circumference  cut  from  the  sphere. 

The  centre  of  the  front  circle  casts  a  shadow  up<?h  this  plane 
whose  perspective  is  c2,  Art.  (280),  c"  being  the  perspective  ot 
the  projection  of  c'  on  this  plane,  and  c"c2  parallel  to  s'r.  the  per- 
spective of  the  projection  of  the  ray  through  c'  on  this  plane. 
<y%  parallel  to  c'm'  is  the  perspective  of  the  shadow  cast  by  the 
radius  c'm'  on  this  plane,  Art.  (267),  and  mjc2  the  perspective  Oi 
the  arc  of  shadow  intersecting  yxx#x  in  xit  a  point  of  the  per- 
spective of  the  shadow. 

Otherwise  thus:  Draw  k'k"  parallel  to  «'»•„  it  may  be  taken  as 
the  vertical  trace  of  a  plane  of  rays  perpendicular  to  V.  This 
plane  cuts  from  the  quarter  sphere  a  semi-circumference.  A  ray 
of  light  through  k'  will  intersect  this  semi-circumference  in  a 
point  of  the  curve  of  shadow.  Revolve  the  plane  about  k'k"  un- 
til it  coincides  with  V.  k'x'k"  will  be  the  revolved  position  of 
the  semi-circumference.  Revolve  the  ray  of  light  through  the 
point  of  sight  about  s'r{  until  it  coincides  with  V ;  s"rx  will  be  its 
revolved  position,  s's"  being  equal  to  s'dxi  Art.  (266).  Through  k' 
draw  k'x"  parallel  to  s"rx,  it  will  be  the  revolved  position  of  the 
ray  through  k',  and  x"  will  be  the  revolved  position  of  a  point  of 
the  shadow.  Through  x'  draw  x's' ;  it  is  the  perspective  of  the 
perpendicular  x'x"  intersecting  k'r{  in  a?w  the  perspective  of  the 
shadow  ca>t  by  k'. 

The  perspective  of  this  curve  of  shadow  evidently  begins  at  e', 
the  point  of  tangency  of  a  line  parallel  to  s'r:.* 

The  point  at  which  the  curve  of  shadow  passes  from  the  spherical 
to  the  cylindrical  part  is  evidently  the  point  in  which  the  intersec- 
tion of  the  plane  of  the  curve  of  shadow  with  the  plane  of  the 
upper  base  meets  the  circumference  of  the  upper  base.  To  de- 
termine the  perspective  of  this  point,  through  #s  draw  xuu{  paral- 
lel to  c'e' ;  it  is  the  perspective  of  a  line  of  the  plane  of  the  curve 
of  shadow  as  also  of  the  plane  of  the  circle  of  which  m^  is  the 
perspective.  ux  is  the  perspective  of  the  point  where  this  line 
pierces  the  plane   of  the  upper  base,  and  c'ux  the  perspective  oi 


LINEAR    PERSPECTIVE.  177 

the  intersection  of  the  plane  of  the  circle  of  shadow  with  the 
plane  of  the  upper  base,  and  p{  the  perspective  of  the  required 
point. 


288.  Problem  76.  To  construct  the  perspective  of  a  sphere  with 
its  shade,  and  shadow  on  the  horizontal  plane. 

Let  the  plane  of  the  picture  be  taken  perpendicular  to  the  right 
line,  joining  the  point  of  sight  with  the  centre  of  the  sphere, 
and  tangent  to  the  sphere;  then,  Fig.  124,  c  will  be  the  horizontal 
and  sr  the  vertical  projection  of  the  centre,  ca  being  equal  to  the 
radius  and  b'c"c"'  the  vertical  projection  of  the  sphere. 

The  perspective  of  the  sphere  will  be  determined  by  a  visual  cone 
tangent  to  it,  and  the  circle  in  which  this  cone  is  intersected  by 
the  plane  of  the  picture  will  be  the  outer  line  of  the  perspective, 
Art.  (274). 

To  construct  this,  pass  a  plane  through  S  and  C  perpendicular 
to  AB.  It  intersects  the  sphere  in  a  great  circle,  and  the  cone  in 
rectilinear  elements  tangent  to  this  circle  and  piercing  V  in  points 
of  the  required  circumference.  amx  is  the  vertical  trace  of  this 
auxiliary  plane.  Revolve  the  plane  about  amt  as  an  axis  until  it 
coincides  with  Y  ;  S,  in  revolved  position,  will  be  at  dx  and  C  at 
c",  and  m's'  described  with  the  centre  c'  and  radius  c"s'  will  be 
the  revolved  position  of  an  arc  of  the  circle  cut  from  the  sphere, 
and  m'dx  that  of  one  of  the  tangents.  In  true  position  this  tan- 
gent pierces  V  at  «i„  and  the  circle  m(u<vx  described  with  s'  as  a 
centre  and  s'm,  as  a  radius  will  be  the  perspective  of  the  sphere. 

The  curve  of  shade  on  the  sphere,  the  line  of  contact  of  a  tan- 
gent cylinder  of  rays,  is  the  circumference  of  a  great  circle  whose 
plane  is  perpendicular  to  the  axis,  Art.  (121),  or  to  the  ray  of 
light. 

The  plane  of  rays  through  the  point,  of  sight  whose  vertical 

trace  is  *'r„  is  perpendicular  to  the  plane  of  the  circle  of  shade, 

and  is  evidently  a  principal  plane,  Art.  (206),  of  the  visual  cone 

by  which  the  perspective  of  this  curve  is  determined,  and  since 

12 


ITS  DESCRIPTIVE   GEOMETRY. 

the  plane  of  the  picture  is  perpendicular  to  this  plane,  its  inter- 
section with  the  cone  will  be  an  ellipse,  all  the  chords  of  which 
perpendicular  to  s'r,  are  bisected  by  it.  This  line  will  then  be  an 
indefinite  axis  of  the  ellipse.  Ana.  Geo^  Art.  (85).  This  princi- 
pal plane  intersects  the  circle  of  shade  in  a  diameter  whose  per- 
spective will  be  this  axis. 

To  determine  it,  revolve  the  plane  about  s'rx  until  it  coincides 
with  V.  S  is  found  at  s",  s's"  being  equal  to  8fdx.  The  centre 
of  the  circle  cut  from  the  sphere  will  be  at  c"\  n'l'p'  will  be  the 
revolved  position  of  the  circle,  and  n'p'  perpendicular  to  s''i\  will 
be  the  revolved  position  of  the  diameter.  Two  visual  rays  from 
its  extremities  pierce  V  at  n^  and  p{)  and  nypx  is  its  perspective 
and  the  required  axis. 

o„  the  middle  point  of  this  axis,  is  the  centre  of  the  ellipse, 
and  l[kl  perpendicular  to  nxp^  the  position  of  the  other  axis,  and 
o'  the  revolved  position  of  the  point  of  which  o,  is  the  perspective. 
If  through  IJc^  and  S  a  plane  be  passed,  it  will  intersect  the  circle 
of  shade  in  a  chord  passing  through  the  point  of  wrhich  o'  is  the 
revolved  position,  perpendicular  to  the  plane  of  rays  of  which  sV, 
is  the  trace  and  equal  to  l'kr.  The  perspective  of  this  chord  will 
be  the  other  axis* 

To  determine  it,  revolve  this  plane  about  k{lx  until  it  coincides 
with  V.  S  will  be  found  at  s'",  o/'r  being  equal  to  o,s"  ;  o'  will 
be  at  o",  oxo"  being  equal  to  Ojo',  and  k"l"  perpendicular  to  s'r[ 
will  be  the  revolved  position  of  the  chord,  the  perspective  of 
which  is  k{llt  the  other  axis. 

The  ellipse  described  upon  these  two  axes  is  the  perspective  of  the 
curve  of  shade.  It  is  tangent  to  mxuxvx  at  ux  and  vu  since  these 
are  the  perspectives  of  two  points  which  are  determined  by  two 
visual  planes  of  rays  tangent  to  the  sphere,  whose  traces  are  r,w, 
and  rxvv  These  are  the  perspectives  of  the  points  in  which  the 
curve  of  shade  crosses  the  apparent  contour  of  the  sphere,  Art. 
(275). 

The  curve  of  shadow  on  the  horizontal  'plane  is  cast  by  the  curve 
of  shade*     c2  is  the  perspective  of  the  shadow  cast  by  the  centre, 


LINEAR    PERSPECTIVE.  179 

cx  being  the  perspective  of  its  horizontal  projection,  and  c  r'  the 
perspective  of  the  horizontal  projection  of  a  ray  through  the 
centre,  Art.  (280). 

The  perspective  of  the  shadow  of  each  diameter  of  the  curve 
of  shade  must  pass  through  c2.  If,  then,  we  find  the  perspective 
of  the  point  in  which  a  diameter  pierces  the  horizontal  plane  and 
join  it  with  c2,  we  shall  have  the  indefinite  perspective  of  the 
shadow  of  this  diameter,  and  the  perspectives  of  two  rays  through 
its  extremities  will  intersect  the  perspective  of  this  shadow  in  points 
of  the  perspective  of  the  curve  of  shadow. 

All  diameters  of  the  circle  of  shade  pierce  the  horizontal  plane 
in  the  horizontal  trace  of  the  plane  of  this  circle.  This  trace 
being  a  horizontal  line  must  vanish  in  the  horizon  of  the  picture 
s'd  ,  Art.  (267),  and  being  in  the  plane  of  the  circle  of  shade  must 
also  vanish  in  the  vanishing  line  of  this  plane ;  and  hence  the  in- 
tersection of  these  two  lines  will  be  a  point  of  the  perspective  of 
the  horizontal  trace. 

To  construct  the  vanishing  line  of  the  plane  of  shade,  through 
s"  draw  s"x'  perpendicular  to  s'V, ;  it  is  the  revolved  position  of  a 
line  passing  through  S  in  the  plane  of  rays  whose  vertical  trace  is 
r,s'  and  parallel  to  the  plane  of  shade.  It  pierces  V  at  x\  one 
point  of  the  vertical  trace  of  a  plane  through  S  parallel  to  the 
circle  of  shade.  Through  x  draw  x'qx  perpendicular  to  jtV„  Art. 
(43) ;  it  is  the  vertical  trace,  or  required  vanishing  line.  This  in- 
tersects s'qx  at  <?,. 

To  determine  another  point  of  the  perspective  of  the  horizontal 
trace,  find  the  perspective  of  the  point  in  which  that  diameter  of 
the  circle  of  shade,  which  is  parallel  to  the  plane  of  the  picture, 
pierces  H.  The  horizontal  projection  of  this  diameter  is  parallel 
to  AB;  hence,  c{ix  is  its  perspective,  s'a'1  perpendicular  to  r,s'  is 
the  perspective  of  the  diameter  itself,  for  these  two  •  lines'  are  the 
orthographic  projections  of  the  diameter  and  ray  of  light  through 
the  centre,  since  the  visual  plane  through  each  of  them  contains 
the  projecting  perpendicular  Ss',  Art.  (36).    ix  is  then  the  perspec- 


ISO  DKSCRIPTIVK    GEOMETRY. 

tive  of  the  point  in  which  this  diameter  pierces  II,  and  qxi{  the 
perspective  of  the  horizontal  trace. 

Join  ix  with  r2;  «,<?,  is  the  perspective  of  the  shadow  cast  by  the 
diameter  represented  by  g f\.  Join  g,  and/,  with  r,  by  right  lines. 
These  are  the  perspectives  of  rays  through  the  extremities  of 
the  diameter,  and  g2  and  /a  are  the  perspectives  of  points  of 
the  curve  of  shadow. 

Produce  w,«,  to  y,,  y,  is  the  perspective  of  the  point  in  which 
the  corresponding  diameter  pierces  II,  and  y,ra  the  perspective  of 
its  shadow,  and  ?2  and  u.2  the  perspectives  of  the  shadows  cast  by 
its  extremities.  In  the  same  way  other  points  of  the  curve  are 
constructed. 

The  curve  must  be  drawn  tangent  to  rlul  and  r,v,  at  w2  and  v2, 
since  these  are  the  perspectives  of  the  points  in  which  the  curve 
of  shadow  crosses  the  elements  of  apparent  contour  of  the  cylin- 
der of  rays,  by  which  the  curve  of  shadow  is  determined. 


289.  Problem  77.  To  construct  the  perspective  of  a  groined 
arch  with  its  shadows. 

To  form  the  groined  arch  here  considered,  two  equal  right  semi- 
cylinders,  with  semi-circular  bases,  are  so  placed  that  .their  axes 
shall  be  at  right  angles  and  bisect  each  other  and  the  diameters 
of  the  semi-circular  bases  in  the  same  horizontal  plane.  Thus, 
Fig.  125,  let  nolu  be  the  horizontal,  and  ok' I  the  vertical  projec- 
tion of  one  semi-cylinder,  and  fghi  the  horizontal,  and  pp'q'q  the 
vertical  projection  of  the  second  cylinder,  pp  and  qq'  being  the 
vertical  projections  of  the  two  semi-circular  bases. 

These  two  cylinders  intersect  in  two  equal  ellipses  horizontally 
projected  in  the  diagonals  <ib  and  ce.  These  ellipses  are  the 
groins.  The  arch  is  now  formed  by  taking  out  from  each  cyl- 
inder that  part  of  the  other  which  lies  within  it.  Thus,  all  that 
part  of  the  cylinder  whose  elements  are  parallel  to  V,  horizontally 
projected  in  akc  and  bke,  is  removed,  as  also  that  part  of  the  othoi 
cylinder  projected  in  ake  and  bkc. 


LINEAR   PERSPECTIVE.  181 

The  arch  thus  formed  is  placed  upon  four  pillars  standing  in 
the  four  corners  of  a  square,  and  whose  horizontal  projections  are 
the  four  squares  mnai,  uchv,  etc.  The  elements  ih  and  fg  being 
coincident  with  the  inner  upper  edges  of  the  pillars  which  are 
horizontally  projected  in  ia-ch  and  fe-bg1  and  the  elements  no  and 
ul  coincident  with  the  inner  upper  edges  which  are  projected  in 
na-eo  aud  uc-bl. 

In  this  position  of  the  arch  the  front  circumference,  or  front 
base  of  the  cylinder  whose  elements  are  perpendicular  to  V, 
springs  from  the  upper  extremities  of  the  two  edges  of  the  front 
pillars  which  are  horizontally  projected  at  n  and  u  and  is 
described  on  a  diameter  equal  and  parallel  to  nu.  The  back 
circumference,  or  other  base  of  the  same  cylinder,  springs  from 
the  upper  extremities  of  the  edges  horizontally  projected  at  ^> 
and  I. 

The  side  circumferences,  or  bases  of  the  other  cylinder,  spring, 
the  one  from  the  upper  ends  of  the  two  edges  horizontally  pro- 
jected in  i  and  /,  and  the  other  from  those  projected  in  h  and  g, 
and  their  diameters  are  parallel  and  equal  to  if  and  kg. 

The  groins  spring  from  the  upper  ends  of  the  four  inner  edges 
of  the  pillars,  the  one  from  those  projected  in  a  and  b,  and  the 
other  from  those  projected  in  c  and  e,  and  ab  and  ce  are  respect- 
ively equal  and  parallel  to  the  transverse  axes  of  the  groins,  the 
common  conjugate  axis  being  equal  and  parallel  to  the  radius 
h'h". 

The  planes  of  the  outer  faces  of  the  pillars  are  produced  up- 
wards, inclosing  the  mass  of  masonry  supported  by  the  arch. 

To  construct  the  perspective  of  the  arch  thus  placed  let  us  take, 
Fig.  126,  the  plane  of  the  picture  coincident  with  the  plane  of  the 
front  faces  of  the  front  pillars,  mm'n'n  and  uu'v'v  being  these 
front  faces  and  their  own  perspectives,  and  let  the  principal  point 
be  at  s/  in  a  plane  perpendicular  to  V  and  midway  between  the 
pillars. 

The  perspectives  of  the  front  pillars  are  constructed  precisely  a& . 
in  Art.  (282)  ;  a{  and  ix  being  the  perspectives  of  the  upper  ends 


182  DESCRIPTIVE   GEOMETRY. 

of  the  two  back  edges  of  the  first,  and  cx  and  \  those  of  the 
second  pillar ;  mdt,  the  perspective  of  the  diagonal  corresponding 
to  mq  in  Fig.  125,  intersects  us',  the  perspective  of  the  perpendic- 
ular corresponding  to  ul,  in  b.21  which  is  the  perspective  of  the 
lower  end  of  the  edge  of  the  back  pillar  corresponding  to  b  ;  b2g^ 
parallel  to  AB,  is  the  perspective  of  the  edge  corresponding  to  bg, 
and  b2la  that  corrresponding  to  bl.  Through  b2y  g2,  and  1.2  draw 
b2bu  g^g^  and  ljx  until  they  meet  u's'  and  v's'  in  5„  g{,  and  lu  and 
draw  b{g{  and  6,/,,  thus  completing  the  perspectives  of  the  two. 
faces  of  the  back  pillars  which  can  be  seen. 

In  the  same  way  the  perspective  of  the  other  back  pillar  may 
be  constructed,  e2  being  the  perspective  of  the  point  corresponding 
to  e. 

On  n'u'  as  a  diameter  describe  the  semicircle  n'k'u';  it  is  the 
front  semicircle  of  the  arch  in  the  plane  of  the  picture  and  its  own . 
perspective.  The  other  base  of  the  same  cylinder  being  parallel 
to  the  plane  of  the  picture  will  be  in  perspective  a  semicircle,  and 
since  it  springs  from  the  points  of  which  o,  and  Z,  are  the  per- 
spectives, o^,  will  be  the  diameter  and  the  semicircle  on  this  its 
perspective. 

To  obtain  points  of  the  perspectives  of  the  groins  and  side  circles, 
intersect  the  arch  by  horizontal  planes.  Each  plane  will  cut  from 
the  cylinders  rectilinear  elements,  the  intersection  of  which  will 
be  points  of  the  groins ;  and  from  the  side  faces  of  the  arch,  right 
lines  which  will  intersect  the  elements  parallel  to  the  plane  of  the 
picture  in  points  of  the  side  circles. 

Let  m"v"  be  the  trace  of  such  a  plane.  It  intersects  the  cylin- 
der whose  axis  is  perpendicular  to  Y  in  two  elements  which  pierce 
V  at  w'  and  z',  vanish  at  s\  and  are  represented  by  w's'  and  z's'. 
The  same  plane  intersects  the  side  faces  of  the  arch  in  lines  repre- 
sented by  m"s'  and  v"s'.  The  two  diagonals  through  m"  and  v", 
intersect  the  elements  in  points  of  the  groins,  see  Fig.  125.  m"d{ 
and  v"dz  are  the  perspectives  of  these  diagonals  intersecting  w's'  \ 
in  w"  and  w"'  and  z's'  in  z"  and  z"\  points  of  the  perspectives  of 
the  groins.     w"z'  and  w'"z'"  are  the  perspectives  of  the  elements 


LINEAR   PERSPECTIVE.  183 

cut  from  the  cylinder  whose  axis  is  parallel  to  V.  These  are  in- 
tersected by  m"s'  and  v"s'  in  points  px  and  q{  of  the  perspective* 
of  the  side  circles. 

The  perspectives  of  the  groins  thus  determined  are  drawn,  one 
from  a,  to  6t,  the  other  from  c,  to  e„  and  the  perspectives  of  the 
side  circles,  one  from  t,  to/,)  and  the  other  from  A,  to  gx. 

Since  all  four  of  these  curves  cross  the  element  of  contour  of 
the  cylinder  whose  axis  is  parallel  to  the  plane  of  the  picture,  their 
perspectives  will  be  tangent  to  the  perspective  of  this  element. 
To  determine  it,  through  S  pass  a  plane  perpendicular  to  the  axis 
of  the  cylinder,  r2s'  is  its  trace.  It  cuts  from  the  cylinder  a  circle 
and  from  the  visual  plane  tangent  to  the  cylinder  a  right  line  tan- 
gent to  the  circle.  Revolve  this  plane  about  r^s'  until  it  coincides 
with  V.  The  centre  of  the  circle  will  be  at  ra',  and  S  at  tf ,.  The 
semicircle  described  with  m'  as  a  centre  and  k"n'  as  a  radius  is 
the  revolved  position  of  the  circle  cut  from  the  cylinder,  and  the 
tangent  to  it  from  dx  the  revolved  position  of  the  line  cut  from 
the  tangent  plane.  This  pierces  V  at  x',  a  point  of  the  perspec- 
tive of  the  element  of  contour.  The  line  through  x'  parallel  to 
AB,  is  the  tangent  to  all  the  curves. 

nr'  and  i2r'  are  the  perspectives  of  the  indefinite  shadows  cast 
on  H  by  the  vertical  edges  of  the  left  hand  pillar,  Art.  (282).  The 
point  in  which  nr'  intersects  btfa  is  the  perspective  of  a  point  of 
the  shadow  of  nn'  on  the  front  face  of  the  back  pillar.  This  line 
being  parallel  to  this  face,  its  shadow  will  be  parallel  to  itself  as 
also  the  perspective  of  the  shadow,  which  is  terminated  at  w2,  the 
perspective  of  the  shadow  cast  by  n',  Art.  (280). 

From  this  point  the  shadow  is  cast  on  this  face  by  the  front 
circle,  and  is  the  arc  of  a  circle  as  is  its  perspective, .  Art. 
(272).  The  shadow  of  the  radius  n'k"  is  parallel  to  itself,  as  also 
the  perspective  of  this  shadow  passing  through  na  and  termi- 
nated at  &2-  The  arc  described  with  Jca  as  a  centre  and  k2n3  as  a 
radius  is  then  the  perspective  of  the  shadow  of  the  front  circle  on 
the  face, 

e%r'  is  the  perspective  of  the  shadow  cast  on  H  by  the  edge  rep 


184  DESCRIPTIVE  GEOMETRY. 

resented  by  e8e„  a  small  part  of  which  only  lies  in  the  limits  of 
the  drawing. 

Points  of  the  shadow  cast  by  the  front  circle  on  the  cylinder,  of 
which  it  is  the  base,  are  determined  by  intersecting  the  cylinder 
by  planes  of  rays  perpendicular  to  the  plane  of  the  picture.  The 
traces  of  these  planes  are  parallel  to  s'rXi  Art.  (279),  and  each 
plane  cuts  the  front  circumference  in  a  point  casting  the  shadow 
and  the  cylinder  in  a  rectilinear  element,  which  is  intersected  by 
the  ray  through  the  point  in  its  shadow. 

Let  y'y"  be  the  trace  of  such  a  plane.  It  intersects  the  cylin- 
der in  an  element  represented  by  y'V,  and  the  circumference  in 
the  point  y1,  and  y2  is  the  perspective  of  the  shadow.  The  per- 
spective of  the  shadow  begins  at  the  point  in  which  a  trace  par- 
allel to  y"y'  is  tangent  to  the  circle. 

Points  of  the  shadow  cast  by  the  side  circle  on  the  left,  on  the 
cylinder  of  which  it  is  a  base,  may  be  found  by  intersecting  the 
cylinder  by  planes  of  rays  perpendicular  to  the  side  faces  of  the 
arch.  Each  plane  intersects  the  circumference  in  a  point  casting 
the  shadow  and  the  cylinder  in  a  rectilinear  element  which  is  in- 
tersected by  the  ray  through  the  point  in  a  point  of  the  shadow. 

The  intersections  of  these  planes  with  the  outer  side  face  of  the 
arch  vanish  at  rw  Art.  (278).  Let  r2£„  be  the  perspective  of  one 
of  these  intersections.  The  plane  intersects  the  side  circumference 
in  a  point  of  which  tx  is  the  perspective  and  the  cylinder  in  an 
element  represented  by  pxq^  and  t2  is  the  perspective  of  the  point 
of  shadow.  The  perspective  of  the  curve  evidently  begins  at  the 
point  of  tangency  of  a  line  through  r2  tangent  to  ixp\f\. 


290.  Although  in  general  the  perspectives  of  objects  are  made 
as  in  the  preceding  articles  on  a  plane,  it  is  not  unusual  to  make 
them  upon  curved  surfaces. 

Extended  panoramic  views  are  thus  made  upon  a  right  cylinder 
with  a  circular  base,  the  point  of  sight  being  in  the  axis,  and  the 
observer  thus  entirely  surrounded  by  pictures  of  objects. 


LINEAR   PERSPECTIVE.  185 

Objects  are  also  sometimes  represented  on  the  interior  of  a 
spherical  dome. 

In  all  cases  the  perspectives  of  points  are  constructed  by  draw- 
ing through  the  points  visual  rays,  and  finding  the  points  in  which 
these  pierce  the  surface  on  which  the  representation  is  made. 

The  constructions  involve  only  the  principles  contained  in  Arts. 
(263),  (264),  and  (268). 


PART  V. 

ISOMETRIC    PROJECTIONS. 


PRELIMINARY    DEFINITIONS    AND    PRINCIPLES. 

291.  Let  three  right  lines  be  drawn  intersecting  in  a  common 
point  and  perpendicular  to  each  other,  two  of  them  being  hori- 
zontal and  the  third  vertical ;  as  the  three  rectangular  co-ordinate 
axes  in  space,  Ana.  Geo.,  Art.  (42),'  or  the  three  adjacent  edges  of 
a  cube;  then  let  a  fourth  right  line  be  drawn  through  the  same 
point,  making  equal  angles  with  the  first  three,  as  the  diagonal  of 
a  cube.  If  a  plane  be  now  passed  perpendicular  to  this  fourth 
line,  and  the  right  lines  and  other  objects  be  orthographically  pro- 
jected upon  it,  the  projections  are  called  Isometric. 

The  three  right  lines  first  drawn  are  the  co-ordinate  axes ;  and 
the  planes  of  these  axes,  taken  two  and  two,  are  the  co-ordinate 
planes.     The  common  point  is  the  origin. 

The  fourth  right  line  is  the  Isometric  axis. 

If  A  designate  the  origin,  the  co-ordinate  axes  are  designated, 
as  in  Ana.  Geo.,  as  the  axes  AX,  AY,  and  AZ,  the  latter  being 
vertical ;  and  the  co-ordinates  planes  as  the  plane  XY,  XZ  and 
YZ,  the  first  being  horizontal,  and  the  other  two  vertical. 


292.  Since  the  co-ordinate  axes  make  equal  angles  with  each 
other  and  with  the  plane  of  projection,  it  is  evident  that  their  pro- 


ISOMETRIC    PROJECTIONS.  187 

jections  will  make  equal  angles  with  each  other,  two  and  two,  that 
is,  angles  of  120°.  Hence,  Fig.  128,  if  any  three  right  lines,  as 
Ax,  Ay,  and  As,  be  drawn  through  a  point  as  A,  making  with 
each  other  angles  of  120°,  these  may  be  taken  as  the  projections 
of  the  co-ordinate  axes,  and  are  the  directrices  of  the  drawing. 

It  is  further  evident,  that  if  any  equal  distances  be  taken  on 
the  co-ordinate  axes,  or  on  lines  parallel  to  either  of  them,  their 
projections  will  be  equal  to  each  other,  since  each  projection  will 
•  be  equal  to  the  distance  itself  into  the  cosine  of  the  angle  of  in- 
clination of  the  axes  with  the  plane  of  projection,  Art.  (197). 

The  angle  which  the  diagonal  of  a  cube  makes  with  either  ad- 
jacent edge  is  known  to  be  54°  44' ;  therefore  the  angle  which 
either  edge,  or  either  of  the  co-ordinate  axes,  makes  with  the 
plane  of  projection  will  be  the  complement  of  this  angle,  viz., 
35°  16'. 


293.  If  a  scale  of  equal  parts,  Fig.  127,  be  constructed,  the  unit 
of  the  scale  being  the  projection  of  any  definite  part  of  either  co- 
ordinate axis,  as  one  inch,  one  foot,  etc.,  that  is,  one  inch  multiplied 
by  the  natural  cosine  of  35°  1^'  =  . 8164V  of  an  inch;  we  may, 
from  this  scale,  determine  the  true  length  of  the  isometric  projec- 
tion of  any  given  portion  of  either  of  the  co-ordinate  axes,  or  of 
lines  parallel  to  them,  by  taking  from  the  scale  the  same  number 
of  units  as  the  number  of  inches,  or  feet,  etc.,  in  the  given  distance. 
Conversely,  the  true  length  of  any  corresponding  line  in  space  may 
be  found  by  applying  its  projection  to  the  isometric  scale,  and 
taking  the  same  number  of  inches,  or  feet,  etc.,  as  the  number  oi 
parts  covered  on  the  scale. 


294.  Since  in  most  of  the  frame  work  connected  with  ma- 
chinery, and  in  the  various  kinds  of  buildings,  the  principal  lines 
to  be  represented  occupy  a  position  similar  to  that  of  the  co-ordi- 
nate axes,  viz.,  perpendicular  to  each  other,  one  system  being  ver- 


188  DESCRIPTIVE    GEOMETRY. 

tical,  and  two  others  horizontal,  the  isometric  projection  is  used  to 
great  advantage  in  their  representation. 

A  still  greater  advantage  arises  from  the  fact  that  in  a  drawing 
thus  made,  all  lines  parallel  to  the  directrices  are  constructed  on 
the  same  scale,  Art.  (292). 


ISOMETRIC    PROJECTIONS    OF    POINTS    AND    LINES. 

295.  If  a  point  be  given,  as  in  Analytical  Geometry,  by  its 
co-ordinates,  or  its  three  distances  from  the  co-ordinate  planes, 
Art.  (40),  its  isometric  projection  may  be  easily  constructed. 
Thus,  Fig.  128,  let  Ax,  Ay,  and  Az  be  the  directrices,  A  being  the 
projection  of  the  origin.  On  Ax  lay  ofF  Apt  equal  to  the  same 
number  of  units,  taken  from  the  isometric  scale,  as  there  are  units 
in  the  distance  of  the  point  from  the  co-ordinate  plane  YZ. 
Through  p  draw  pm'  parallel  to  Ay,  and  make  it  equal  to  the 
number  of  units  in  the  distance  of  the  point  from  the  plane  XZ. 
Through  m!  draw  m'm  parallel  to  Az,  and  make  it  equal  to  the 
third  given  distance,  and  m  will  be  the  required  projection. 


296.  The  projection  of  any  right  line  parallel  to  either  of  the 
co-ordinate  axes  m  ty  be  constructed  by  finding,  as  above,  the  pro- 
jection of  one  of  its  points,  and  drawing  through  this  a  line 
parallel  to  the  proper  directrix. 

If  the  line  is  parallel  to  neither  of  the  axes,  the  proj  .^tions  of 
two  of  its  points  may  be  found  as  above  and  joined  ha  right 
line. 

The  projections  of  curves  may  be  constructed  by  finding  a 
sufficient  number  of  the  projections  of  their  points. 


297.  If  the  circumference  of  a  circle  be  in,  or  parallel  to,  either 
co-ordinate  plane,  its  projection  may  be  constructed  thus :  At  the 


ISOMETRIC    PROJECTIONS.  189 

extremities  of  the  two  diameters,  which  are  parallel  to  the  co  or- 
dinate axes,  draw  tangents,  thus  circumscribing  the  circle  by  a 
square.  The  projections  of  each  set  of  tangents  will  be  parallel  to 
the  projection  of  the  parallel  diameter  and  tangent  to  the  projec- 
tion of  the  circle,  Art.  (65)  ;  hence  the  projections  of  these  two 
equal  diameters  will  be  equal  conjugate  diameters  of  the  ellipse 
which  is  the  projection  of  the  circle,  and  since  these  projections 
are  parallel  to  the  directrices  they  will  make  with  each  other  an 
angle  of  120°.  Upon  these  the  ellipse  may  be  described,  taking 
care  to  make  it  tangent  to  the  projections  of  the  four  tangents. 


PRACTICAL    PROBLEMS. 

298.  Problem  78.  To  construct  the  isometric  projection  of  a 
tube. 

Let  the  origin  be  taken  at  one  of  the  upper  corners  of  the  cube, 
the  base  being  horizontal,  and  let  Ax,  Ay,  and  Az,  Fig.  129,  be 
the  directrices. 

From  A,  on  the  directrices,  lay  off  the  distances  Ax,  Ay,  and 
Az,  each  equal  to  the  same  number  of  units,  taken  from  the 
isometric  scale,  as  there  are  units  of  length  in  the  edge  of  the 
cube.  These  will  be  the  projections  of  the  three  edges  of  the 
cube  which  intersect  at  A.  Through  x,  y,  and  z  draw  xe,  xg,  ye, 
yc,  zc,  and  zg  parallel  to  the  directrices,  completing  the  three 
equal  rhombuses,  Axey,  etc.  These  will  be  the  projections  of  the 
three  faces  of  the  cube  which  are  seen,  and  the  representation  will 
be  complete. 


299.  The  ellipses  constructed  upon  the  equal  lines  hi,  qs ;  si 
uv,  &c,  Fig.  129,  are  evidently  the  projections  of  three  circles  in- 
scribed in  the  squares  which  form  the  faces  of  the  cube,  and  these 
lines  are  the  equal  conjugate  diameters  of  the  ellipses,  Art.  (297) 
Ik  being  drawn  through  the  middle  point  of  Ax,  and  sq  througi 


190  DESCRIPTIVE   GEOMETRY. 

the  middle  point  of  Az,  and  parallel  respectively  to  Az  and  Ax, 
m  being  the  projection  of  the  centre  of  the  circle. 


800.  Problem  *79.  To  construct  the  isometric  projection  of  an 
upright  rectangular  beam  with  its  shade  and  shadow  on  the  hori- 
zontal plane. 

Let  the  origin  A  and  the  directrices,  Fig.  130,  be  taken  as  in 
the  preceding  problem. 

On  Ax  lay  off  the  distance  Ax  equal  to  the  breadth  of  the  beam 
in  units  taken  from  the  isometric  scale ;  on  Ay,  its  thickness,  and 
on  Az,  its  length,  and  complete  the  three  parallelograms  Axcz, 
Aybz,  and  Axey,  These  will  be  the  projections  of  the  three  faces 
of  the  beam  which  are  seen. 

Let  a'  be  assumed  as  the  isometric  projection  of  the  point  in 
which  a  ray  of  light  through  A  pierces  H ;  then  Aa'  will  be  the 
isometric  projection  of  the  ray,  and  za!  the  projection  of  the 
shadow  of  the  edge  AZ,  Art.  (248).  Through  a!  draw  a'y'  paral- 
lel and  equal  to  Ay.  It  is  the  projection  of  the  shadow  of  the 
edge  AY.  y'e'  equal  and  parallel  to  ye  is  the  projection  of  the 
shadow  cast  by  YE,  and  de'  that  of  the.  edge  DE. 

The  face  represented  by  Azby  is  in  the  shade. 


301.  Problem  80.  To  construct  the  isometric  projection  of  the 
framework  of  a  simple  horizontal  platform  resting  on  four  rectan- 
gular supports. 

Let  A,  Fig.  131,  be  the  upper  corner  of  one  of  the  horizontal 
beams,  the  directrices  being  as  in  the  preceding  problems. 

Lay  off  A#,  A6,  and  Az  equal  respectively  to  the  number  of 
units  in  the  length,  breadth,  and  thickness  of  the  horizontal  beam, 
and  complete  the  three  parallelograms  Ay,  Ax\  Aft',  thus  forming 
the  representation  of  the  beam.  Lay  off  zc  equal  to  the  distance 
of  the  first  support  from  the  end  of  the  beam,  cd  equal  to  its 
thickness,  cc'  parallel  to  As  equal  to  its  height,  and  c'e'  parallel  to 


ISOMBTRIC   PROJECTIONS.  191 

Ay  its  breadth,  and  complete  the  parallelograms  dc'  and  ce'.  Lay 
off  zf  equal  to  the  distance  of  the  second  support  from  the  end  ot 
the  beam,  and  complete  its  projection  in  the  same  way. 

Lay  off  ba  equal  to  the  horizontal  distance  between  the  two 
beams,  and  construct  the  projection  of  the  second  beam  and  its 
supports  precisely  as  the  first  was  constructed. 

From  a  to  n  lay  off  the  distance  from  the  end  of  the  beam  to 
the  first  cross-piece.  Make  nn'  equal  to  its  breadth,  and  np  paral- 
lel to  Az  equal  to  its  thickness,  and  draw  lines  through  n,  ri,  and 
p  parallel  to  Ay.  Lay  off  ar  equal  to  the  distance  of  the  second 
cross-piece  from  the  end  of  the  beam,  and  construct  its  projection 
in  the  same  way  as  the  first. 

The  faces  of  the  frame-work  in  the  shade  and  seen  are  darkened 
in  the  drawing. 


302.  Problem  81.  To  construct  the  isometric  projection  of  a 
house  with  a  projecting  roof. 

Let  A,  Fig.  132,  be  the  upper  end  of  the  intersection  of 
the  front  and  side  faces  of  the  house,  Ax,  Ay,  and  Az  the  direc- 
trices. « 

On  Ax  lay  off  the  length,  on  Ay  the  breadth  of  the  house,  and 
on  Az  the  height  of  the  side  walls,  and  complete  the  two  paral- 
lelograms Ap  and  Ap',  they  are  the  projections  of  the  side  and 
front  faces  of  the  house. 

At  6,  the  middle  point  of  Ay,  draw  br  parallel  to  As,  and  make 
it  equal  to  the  height  of  the  ridge  of  the  roof  above  AY,  and  join^ 
rA  and  ry  ;  these  lines  will  be  the  projections  of  the  intersections 
of  the  face  ZY  with  the  roof.  From  r  lay  off  rr'  equal  to  the  dis- 
tance that  the  roof  projects  beyond  the  walls  of  the  house,  and 
draw  r's  and  r's'  parallel  respectively  to  rA  and  ry.  On  Ay  lay 
off  Ad  and  ye  each  equal  to  rr',  and  draw  dd'  and  ee'  parallel  to 
Az,  intersecting  Ar  and  ry  produced  in  d'  and  e'.  These  will  be 
points  of  the  projections  of  the  eaves.  Draw  r'r"  parallel  to  Ax, 
and  make  it  equal  to  the  length  of  the  ridge,  and  complete  tho 


192  DESCRIPTIVE   GEOMETRY. 

parallelograms  r't  and  r't'.  These  will  be  the  projections  of  the 
two  inclined  faces  of  the  roof. 

From  b  lay  off  be  and  bf,  each  equal  to  the  distance  of  the  side 
faces  of  the  chimney  from  the  ridge,  and  draw  eg  and  fg'  parallel 
to  Az,  intersecting  Ar  and  ry  in  g  and  g\  and  draw  gh  and  g'h! 
parallel  to  r'r".  They  will  be  the  projections  of  the  intersections 
of  the  planes  of  the  side  faces  of  the  chimney  with  the  roof. 
Make  gh  equal  to  the  distance  of  the  front  face  of  the  chimney 
from  the  front  face  of  the  house,  and  draw  hi  and  ih'  parallel  to 
Ar  and  ry;  they  will  be  the  projections  of  the  intersection  of  the 
front  face  of  the  chimney  with  the  roof.  Make  hk  equal  to  the 
thickness  of  the  chimney,  and  hi  equal  to  its  height,  and  complete 
its  projection  as  in  the  Figure. 

From  z  lay  off  zu  equal  to  the  distance  of  the  door  from  the 
edge  Az,  uu'  equal  to  its  width,  and  uw  equal  to  its  height,  and 
complete  the  parallelogram  uw.  Make  zq  equal  to  the  distance 
of  the  windows  above  the  base,  qn  and  qo  their  distances  from  the 
edge,  Az,  nn\  and  00'  their  width,  and  nm  equal  to  their  height, 
and  complete  the  parallelograms. 


303.  A  knowledge  of  the  preceding  simple  principles  and  con- 
structions will  enable  the  draughtsman  to  make  isometric  draw- 
ings of  the  most  complicated  pieces  of  machinery,  and  most  ex- 
tended collections  of  buildings,  walls,  etc. ;  drawings  which  not 
only  present  to  the  eye  of  the  observer  a  very  good  representa- 
tion of  the  objects  projected,  but  are  of  great  use  to  the  machinist 
and  builder. 


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